Class XII Chapter 7 – Integrals Mathscbseocean.weebly.com/uploads/2/8/1/5/28152469/cmn7.pdfClass...

Class XII Chapter 7 – Integrals Maths

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Exercise 7.1

Question 1:

sin 2x

Answer

The anti derivative of sin 2x is a function of x whose derivative is sin 2x.It is known that,

Therefore, the anti derivative of

Question 2:

Cos 3x

Answer

The anti derivative of cos 3x is a function of x whose derivative is cos 3x.

It is known that,

Therefore, the anti derivative of .

Question 3:

Answer

The anti derivative of e2x is the function of x whose derivative is e2x.

It is known that,

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Question 4:

Answer

The anti derivative of is the function of x whose derivative is .

It is known that,

Question 5:

Answer

The anti derivative of is the function of x whose derivative is

It is known that,

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Therefore, the anti derivative of is .

Question 6:

Answer

Question 7:

Answer

Question 8:

Answer

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Question 9:

Answer

Question 10:

Answer

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Question 11:

Answer

Question 12:

Answer

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Question 13:

Answer

On dividing, we obtain

Question 14:

Answer

Question 15:

Answer

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Question 16:

Answer

Question 17:

Answer

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Question 18:

Answer

Question 19:

Answer

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Question 20:

Answer

Question 21:

The anti derivative of equals

(A) (B)

(C) (D)

Answer

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Hence, the correct Answer is C.

Question 22:

If such that f(2) = 0, then f(x) is

(A) (B)

(C) (D)

Answer

It is given that,

∴Anti derivative of

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Hence, the correct Answer is A.

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Exercise 7.2

Question 1:

Answer

Let = t

∴2x dx = dt

Question 2:

Answer

Let log |x| = t

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Question 3:

Answer

Let 1 + log x = t

Question 4:

sin x ⋅ sin (cos x)

Answer

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sin x ⋅ sin (cos x)

Let cos x = t

∴ −sin x dx = dt

Question 5:

Answer

∴ 2adx = dt

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Question 6:

Answer

Let ax + b = t

⇒ adx = dt

Question 7:

Answer

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∴ dx = dt

Question 8:

Answer

Let 1 + 2x2 = t

∴ 4xdx = dt

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Question 9:

Answer

∴ (2x + 1)dx = dt

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Question 10:

Answer

Question 11:

Answer

∴ dx = dt

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Question 12:

Answer

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Question 13:

Answer

∴ 9x2 dx = dt

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Question 14:

Answer

Let log x = t

Question 15:

Answer

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∴ −8x dx = dt

Question 16:

Answer

∴ 2dx = dt

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Question 17:

Answer

∴ 2xdx = dt

Question 18:

Answer

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Question 19:

Answer

Dividing numerator and denominator by ex, we obtain

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Question 20:

Answer

Question 21:

Answer

Let 2x − 3 = t

∴ 2dx = dt

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Question 22:

Answer

Let 7 − 4x = t

∴ −4dx = dt

Question 23:

Answer

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Question 24:

Answer

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Question 25:

Answer

Question 26:

Answer

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Question 27:

Answer

Let sin 2x = t

Question 28:

Answer

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∴ cos x dx = dt

Question 29:

cot x log sin x

Answer

Let log sin x = t

Question 30:

Answer

Let 1 + cos x = t

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Question 31:

Answer

Let 1 + cos x = t

Question 32:

Answer

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Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt

Question 33:

Answer

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Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt

Question 34:

Answer

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Question 35:

Answer

Let 1 + log x = t

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Question 36:

Answer

Question 37:

Answer

Let x4 = t

∴ 4x3 dx = dt

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From (1), we obtain

Question 38:

equals

Answer

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Hence, the correct Answer is D.

Question 39:

equals

Answer

Hence, the correct Answer is B.

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Exercise 7.3

Question 1:

Answer

Question 2:

Answer

It is known that,

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Question 3:

cos 2x cos 4x cos 6x

Answer

It is known that,

Question 4:

sin3 (2x + 1)

Answer

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Question 5:

sin3 x cos3 x

Answer

Question 6:

sin x sin 2x sin 3x

Answer

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It is known that,

Question 7:

sin 4x sin 8x

Answer

It is known that,

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Question 8:

Answer

Question 9:

Answer

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Question 10:

sin4 x

Answer

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Question 11:

cos4 2x

Answer

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Question 12:

Answer

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Question 13:

Answer

Question 14:

Answer

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Question 15:

Answer

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Question 16:

Answer

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From equation (1), we obtain

Question 17:

Answer

Question 18:

Answer

Question 19:

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Answer

Question 20:

Answer

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Question 21:

sin−1 (cos x)

Answer

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It is known that,

Substituting in equation (1), we obtain

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Question 22:

Answer

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Question 23:

is equal to

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

Answer

Question 24:

equals

A. − cot (exx) + C

B. tan (xex) + C

C. tan (ex) + C

D. cot (ex) + C

Answer

Let exx = t

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Exercise 7.4

Question 1:

Answer

Let x3 = t

∴ 3x2 dx = dt

Question 2:

Answer

Let 2x = t

∴ 2dx = dt

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Question 3:

Answer

Let 2 − x = t

⇒ −dx = dt

Question 4:

Answer

Let 5x = t

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∴ 5dx = dt

Question 5:

Answer

Question 6:

Answer

Let x3 = t

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∴ 3x2 dx = dt

Question 7:

Answer

From (1), we obtain

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Question 8:

Answer

Let x3 = t

⇒ 3x2 dx = dt

Question 9:

Answer

Let tan x = t

∴ sec2x dx = dt

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Question 10:

Answer

Question 11:

Answer

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Question 12:

Answer

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Question 13:

Answer

Question 14:

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Answer

Question 15:

Answer

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Question 16:

Answer

Equating the coefficients of x and constant term on both sides, we obtain

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4A = 4 ⇒ A = 1

A + B = 1 ⇒ B = 0

Let 2x2 + x − 3 = t

∴ (4x + 1) dx = dt

Question 17:

Answer

From (1), we obtain

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Question 18:

Answer

Equating the coefficient of x and constant term on both sides, we obtain

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Substituting equations (2) and (3) in equation (1), we obtain

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Question 19:

Answer

Equating the coefficients of x and constant term, we obtain

2A = 6 ⇒ A = 3

−9A + B = 7 ⇒ B = 34

∴ 6x + 7 = 3 (2x − 9) + 34

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Substituting equations (2) and (3) in (1), we obtain

Question 20:

Answer

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Using equations (2) and (3) in (1), we obtain

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Question 21:

Answer

Let x2 + 2x +3 = t

⇒ (2x + 2) dx =dt

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Question 22:

Answer

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Substituting (2) and (3) in (1), we obtain

Question 23:

Answer

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Question 24:

equals

A. x tan−1 (x + 1) + C

B. tan− 1 (x + 1) + C

C. (x + 1) tan−1 x + C

D. tan−1 x + C

Answer

Question 25:

equals

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Answer

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Exercise 7.5

Question 1:

Answer

A + B = 1

2A + B = 0

On solving, we obtain

A = −1 and B = 2

Question 2:

Answer

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A + B = 0

−3A + 3B = 1

Question 3:

Answer

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

A = 1, B = −5, and C = 4

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Question 4:

Answer

Substituting x = 1, 2, and 3 respectively in equation (1), we obtain

Question 5:

Answer

Substituting x = −1 and −2 in equation (1), we obtain

A = −2 and B = 4

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Question 6:

Answer

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (1 − x2) by x(1 − 2x), we obtain

Substituting x = 0 and in equation (1), we obtain

A = 2 and B = 3

Substituting in equation (1), we obtain

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Question 7:

Answer

Equating the coefficients of x2, x, and constant term, we obtain

A + C = 0

−A + B = 1

−B + C = 0

On solving these equations, we obtain

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Question 8:

Answer

Substituting x = 1, we obtain

Equating the coefficients of x2 and constant term, we obtain

A + C = 0

−2A + 2B + C = 0

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Question 9:

Answer

Substituting x = 1 in equation (1), we obtain

Equating the coefficients of x2 and x, we obtain

A + C = 0

B − 2C = 3

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Question 10:

Answer

Equating the coefficients of x2 and x, we obtain

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Question 11:

Answer

Substituting x = −1, −2, and 2 respectively in equation (1), we obtain

Question 12:

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Answer

It can be seen that the given integrand is not a proper fraction.

Therefore, on dividing (x3 + x + 1) by x2 − 1, we obtain

Substituting x = 1 and −1 in equation (1), we obtain

Question 13:

Answer

Equating the coefficient of x2, x, and constant term, we obtain

A − B = 0

B − C = 0

A + C = 2

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A = 1, B = 1, and C = 1

Question 14:

Answer

Equating the coefficient of x and constant term, we obtain

2A + B = −1 ⇒ B = −7

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Question 15:

Answer

Equating the coefficient of x3, x2, x, and constant term, we obtain

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Question 16:

[Hint: multiply numerator and denominator by xn − 1 and put xn = t]

Answer

Multiplying numerator and denominator by xn − 1, we obtain

Substituting t = 0, −1 in equation (1), we obtain

A = 1 and B = −1

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Question 17:

[Hint: Put sin x = t]

Answer

Substituting t = 2 and then t = 1 in equation (1), we obtain

A = 1 and B = −1

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Question 18:

Answer

Equating the coefficients of x3, x2, x, and constant term, we obtain

A + C = 0

B + D = 4

4A + 3C = 0

4B + 3D = 10

A = 0, B = −2, C = 0, and D = 6

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Question 19:

Answer

Let x2 = t ⇒ 2x dx = dt

Substituting t = −3 and t = −1 in equation (1), we obtain

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Question 20:

Answer

Multiplying numerator and denominator by x3, we obtain

Let x4 = t ⇒ 4x3dx = dt

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Substituting t = 0 and 1 in (1), we obtain

A = −1 and B = 1

Question 21:

[Hint: Put ex = t]

Answer

Let ex = t ⇒ ex dx = dt

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Substituting t = 1 and t = 0 in equation (1), we obtain

A = −1 and B = 1

Question 22:

Answer

Substituting x = 1 and 2 in (1), we obtain

A = −1 and B = 2

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Question 23:

Answer

A + B = 0

A = 1, B = −1, and C = 0

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Exercise 7.6

Question 1:

x sin x

Answer

Let I =

Taking x as first function and sin x as second function and integrating by parts, we

obtain

Question 2:

Answer

Let I =

Taking x as first function and sin 3x as second function and integrating by parts, we

obtain

Question 3:

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Answer

Taking x2 as first function and ex as second function and integrating by parts, we obtain

Again integrating by parts, we obtain

Question 4:

x logx

Answer

Taking log x as first function and x as second function and integrating by parts, we

obtain

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Question 5:

x log 2x

Answer

Taking log 2x as first function and x as second function and integrating by parts, we

obtain

Question 6:

x2 log x

Answer

Taking log x as first function and x2 as second function and integrating by parts, we

obtain

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Question 7:

Answer

Taking as first function and x as second function and integrating by parts, we

obtain

Question 8:

Answer

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Taking as first function and x as second function and integrating by parts, we

obtain

Question 9:

Answer

Taking cos−1 x as first function and x as second function and integrating by parts, we

obtain

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Question 10:

Answer

Taking as first function and 1 as second function and integrating by parts, we

obtain

Question 11:

Answer

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Taking as first function and as second function and integrating by parts,

we obtain

Question 12:

Answer

Taking x as first function and sec2x as second function and integrating by parts, we

obtain

Question 13:

Answer

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obtain

Question 14:

Answer

obtain

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Question 15:

Answer

Let I = I1 + I2 … (1)

Where, and

Taking log x as first function and x2 as second function and integrating by parts, we

obtain

Taking log x as first function and 1 as second function and integrating by parts, we

obtain

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Question 16:

Answer

It is known that,

Question 17:

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Answer

Let ⇒

It is known that,

Question 18:

Answer

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Let ⇒

It is known that,

Question 19:

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Answer

Also, let ⇒

It is known that,

Question 20:

Answer

Let ⇒

It is known that,

Question 21:

Answer

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Integrating by parts, we obtain

Question 22:

Answer

Let ⇒

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Question 23:

equals

Answer

Also, let ⇒

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Question 24:

equals

Answer

Also, let ⇒

It is known that,

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Exercise 7.7

Question 1:

Answer

Question 2:

Answer

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Question 3:

Answer

Question 4:

Answer

Question 5:

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Answer

Question 6:

Answer

Question 7:

Answer

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Question 8:

Answer

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Question 9:

Answer

Question 10:

is equal to

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Answer

Question 11:

is equal to

Answer

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Exercise 7.8

Question 1:

Answer

It is known that,

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Question 2:

Answer

It is known that,

Question 3:

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Answer

It is known that,

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Question 4:

Answer

It is known that,

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From equations (2) and (3), we obtain

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Question 5:

Answer

It is known that,

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Question 6:

Answer

It is known that,

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Exercise 7.9

Question 1:

Answer

By second fundamental theorem of calculus, we obtain

Question 2:

Answer

Question 3:

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Answer

Question 4:

Answer

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Question 5:

Answer

Question 6:

Answer

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Question 7:

Answer

Question 8:

Answer

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Question 9:

Answer

Question 10:

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Answer

Question 11:

Answer

Question 12:

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Answer

Question 13:

Answer

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Question 14:

Answer

Question 15:

Answer

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Question 16:

Answer

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A = 10 and B = −25

Substituting the value of I1 in (1), we obtain

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Question 17:

Answer

Question 18:

Answer

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Question 19:

Answer

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Question 20:

Answer

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Question 21:

equals

Answer

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Question 22:

equals

Answer

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Exercise 7.10

Question 1:

Answer

When x = 0, t = 1 and when x = 1, t = 2

Question 2:

Answer

Also, let

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Question 3:

Answer

Also, let x = tanθ ⇒ dx = sec2θ dθ

When x = 0, θ = 0 and when x = 1,

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Takingθas first function and sec2θ as second function and integrating by parts, we obtain

Question 4:

Answer

Let x + 2 = t2 ⇒ dx = 2tdt

When x = 0, and when x = 2, t = 2

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Question 5:

Answer

Let cos x = t ⇒ −sinx dx = dt

When x = 0, t = 1 and when

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Question 6:

Answer

Let ⇒ dx = dt

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Question 7:

Answer

Let x + 1 = t ⇒ dx = dt

When x = −1, t = 0 and when x = 1, t = 2

Question 8:

Answer

Let 2x = t ⇒ 2dx = dt

When x = 1, t = 2 and when x = 2, t = 4

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Question 9:

The value of the integral is

Answer

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Let cotθ = t ⇒ −cosec2θ dθ= dt

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Question 10:

A. cos x + x sin x

B. x sin x

C. x cos x

D. sin x + x cos x

Answer

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Exercise 7.11

Question 1:

Answer

Adding (1) and (2), we obtain

Question 2:

Answer

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Question 3:

Answer

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Question 4:

Answer

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Question 5:

Answer

It can be seen that (x + 2) ≤ 0 on [−5, −2] and (x + 2) ≥ 0 on [−2, 5].

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Question 6:

Answer

It can be seen that (x − 5) ≤ 0 on [2, 5] and (x − 5) ≥ 0 on [5, 8].

Question 7:

Answer

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Question 8:

Answer

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Question 9:

Answer

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Question 10:

Answer

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Question 11:

Answer

As sin2 (−x) = (sin (−x))2 = (−sin x)2 = sin2x, therefore, sin2x is an even function.

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It is known that if f(x) is an even function, then

Question 12:

Answer

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Question 13:

Answer

As sin7 (−x) = (sin (−x))7 = (−sin x)7 = −sin7x, therefore, sin2x is an odd function.

It is known that, if f(x) is an odd function, then

Question 14:

Answer

It is known that,

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Question 15:

Answer

Question 16:

Answer

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sin (π − x) = sin x

Let 2x = t ⇒ 2dx = dt

When x = 0, t = 0 and when

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Question 17:

Answer

It is known that,

Question 18:

Answer

It can be seen that, (x − 1) ≤ 0 when 0 ≤ x ≤ 1 and (x − 1) ≥ 0 when 1 ≤ x ≤ 4

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Question 19:

Show that if f and g are defined as and

Answer

Question 20:

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The value of is

Answer

It is known that if f(x) is an even function, then and

if f(x) is an odd function, then

Question 21:

The value of is

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Answer

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Miscellaneous Solutions

Question 1:

Answer

−A + B − C = 0

B + C = 0

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Question 2:

Answer

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Question 3:

[Hint: Put ]

Answer

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Question 4:

Answer

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Question 5:

Answer

On dividing, we obtain

Question 6:

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Answer

A + B = 0

B + C = 5

9A + C = 0

Question 7:

Answer

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Let x − a = t ⇒ dx = dt

Question 8:

Answer

Question 9:

Answer

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Let sin x = t ⇒ cos x dx = dt

Question 10:

Answer

Question 11:

Answer

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Question 12:

Answer

Let x4 = t ⇒ 4x3 dx = dt

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Question 13:

Answer

Let ex = t ⇒ ex dx = dt

Question 14:

Answer

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Equating the coefficients of x3, x2, x, and constant term, we obtain

A + C = 0

B + D = 0

4A + C = 0

4B + D = 1

Question 15:

Answer

= cos3 x × sin x

Let cos x = t ⇒ −sin x dx = dt

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Question 16:

Answer

Question 17:

Answer

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Question 18:

Answer

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Question 19:

Answer

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Question 20:

Answer

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Question 21:

Answer

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Question 22:

Answer

Equating the coefficients of x2, x,and constant term, we obtain

A + C = 1

3A + B + 2C = 1

2A + 2B + C = 1

A = −2, B = 1, and C = 3

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Question 23:

Answer

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Question 24:

Answer

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Question 25:

Answer

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Question 26:

Answer

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When x = 0, t = 0 and

Question 27:

Answer

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When and when

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Question 28:

Answer

When and when

As , therefore, is an even function.

It is known that if f(x) is an even function, then

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Question 29:

Answer

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Question 30:

Answer

Question 31:

Answer

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Question 32:

Answer

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Question 33:

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Answer

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From equations (1), (2), (3), and (4), we obtain

Question 34:

Answer

A + C = 0

A + B = 0

A = −1, C = 1, and B = 1

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Hence, the given result is proved.

Question 35:

Answer

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Question 36:

Answer

Therefore, f (x) is an odd function.

It is known that if f(x) is an odd function, then

Question 37:

Answer

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Question 38:

Answer

Question 39:

Answer

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Let 1 − x2 = t ⇒ −2x dx = dt

Question 40:

Evaluate as a limit of a sum.

Answer

It is known that,

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Question 41:

is equal to

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Answer

Question 42:

is equal to

Answer

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Question 43:

If then is equal to

Answer

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Question 44:

The value of is

C. − 1

Answer

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(ii) The differential of y, denoted by dy , is defined by dxxfdy 0' or

PRACTICE PROBLEMS

LEVEL -I

Find the slope of the tangent to the curve xxy 43 2 at x = 4

Find the slope of the tangent to the curve 𝑦 = 𝑥2 − 3𝑥 + 2 at the point whose x-

coordinate is 3

Find the slope of the normal to the curve 4

,,sin,cos 33 atayax

LEVEL –II

Find the point at which the tangent to the curve 𝑦 = √4𝑥 − 3 − 1 has its slope 2

Find the equation of the tangent line to the curve 722 xxy which is (i)

parallel to the line 092 yx (ii) Perpendicular to the line 13155 xy

Find the point on the curve 03222 xyx at which the tangents are the

parallel to the x-axis.

LEVEL-III

Find the equation of the normal to the curve 623 xxy which are parallel to the

line x + 14y + 4 = 0.

Prove that the curves 2yx and kxy cut at right angleif 18 3 k

Find the equation of the tangent to the curve 23 xy which is parallel to the line

0524 yx

INTEGRATION

INTRODUCTION

IF f(x) is derivative of function g(x), then g(x) is known as antiderivative or integral of

i.e., 𝒅

𝒅𝒙 (g(x)) = f(x) ⇔ ∫ 𝒇(𝒙)𝒅𝒙 = 𝒈(𝒙)

STANDARD SET OF FORMULAS

* Where c is an arbitrary constant.

1. ∫𝒙𝒏 𝒅𝒙 = 𝒙𝒏+𝟏

𝒏+𝟏 + 𝒄 (n -1)

2. ∫𝒅𝒙 = x + c

3. ∫𝟏

𝒙 dx = log |x| + c

4. ∫ 𝒄𝒐𝒔 𝒙 𝒅𝒙 = 𝒔𝒊𝒏 𝒙 + 𝒄

5. ∫ 𝒔𝒊𝒏 𝒙 𝒅𝒙 = − 𝒄𝒐𝒔 𝒙 + 𝒄

6. ∫ 𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = 𝒕𝒂𝒏 𝒙 + 𝒄

7. ∫ 𝒄𝒐𝒔𝒆𝒄𝟐 𝒙 𝒅𝒙 = −𝒄𝒐𝒕 𝒙 + 𝒄

8. ∫ 𝒔𝒆𝒄 𝒙 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒔𝒆𝒄 𝒙 + 𝒄

9. ∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒄𝒐𝒕 𝒙 𝒅𝒙 = −𝒄𝒐𝒔𝒆𝒄 𝒙 + 𝒄

10. ∫𝒆𝒙 𝒅𝒙 = 𝒆𝒙 + 𝒄

11. ∫ 𝒕𝒂𝒏 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 | + 𝒄

12. ∫ 𝒄𝒐𝒕 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒊𝒏 𝒙 | + 𝒄

13. ∫ 𝒔𝒆𝒄 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒔𝒆𝒄 𝒙 + 𝒕𝒂𝒏 𝒙 | + 𝒄

14. ∫ 𝒄𝒐𝒔𝒆𝒄 𝒙 𝒅𝒙 = 𝒍𝒐𝒈 |𝒄𝒐𝒔𝒆𝒄 𝒙 − 𝒄𝒐𝒕 𝒙 | + 𝒄

15. ∫𝟏

√𝟏− 𝒙𝟐 dx = sin -1 x + c

16. ∫𝟏

𝟏+𝒙𝟐 𝒅𝒙 = 𝒕𝒂𝒏−𝟏𝒙 + 𝒄

17. ∫𝟏

𝒙√𝒙𝟐− 𝟏 𝒅𝒙 = 𝐬𝐞𝐜−𝟏 𝒙 + 𝒄

18. ∫𝒂𝒙 𝒅𝒙 = 𝒂𝒙

𝒍𝒐𝒈 𝒂+ 𝒄

19. ∫𝟏

√𝒙 𝒅𝒙 = 𝟐√𝒙 + 𝒄

INTEGRALS OF LINEAR FUNCTIONS

1. ∫(𝒂𝒙 + 𝒃)𝒏 𝒅𝒙 = (𝒂𝒙+𝒃)𝒏+𝟏

(𝒏+𝟏)𝒂+ 𝒄

2. ∫𝟏

𝒂𝒙+𝒃 𝒅𝒙 =

𝒍𝒐𝒈 (𝒂𝒙+𝒃)

𝒂+ 𝒄

3. ∫ 𝒔𝒊𝒏 (𝒂𝒙 + 𝒃 )𝒅𝒙 = −𝒄𝒐𝒔 (𝒂𝒙+𝒃)

𝒂+ 𝒄

In the same way if ax +b comes in the place of x, in the standard set of formulas, then

divide the integral by a

SPECIAL INTEGRALS

1. ∫𝟏

𝒙𝟐−𝒂𝟐 𝒅𝒙 = 𝟏

𝟐𝒂 𝒍𝒐𝒈 |

𝒙−𝒂

𝒙+𝒂| + c

2. ∫𝟏

𝒂𝟐−𝒙𝟐 𝒅𝒙 = 𝟏

𝟐𝒂 𝒍𝒐𝒈 |

𝒂+𝒙

𝒂−𝒙| + 𝒄

4. ∫𝟏

√𝒙𝟐−𝒂𝟐 𝒅𝒙 = 𝒍𝒐𝒈| 𝒙 + √𝒙𝟐 − 𝒂𝟐 | + 𝒄

5. ∫𝟏

√𝒙𝟐+𝒂𝟐 𝒅𝒙 = 𝒍𝒐𝒈 |𝒙 + √𝒙𝟐 + 𝒂𝟐| + c

6. ∫𝟏

√𝒂𝟐−𝒙𝟐 𝒅𝒙 = 𝐬𝐢𝐧−𝟏 (

𝒂) + 𝒄

7. ∫√𝒙𝟐 + 𝒂𝟐 𝒅𝒙 =𝒙

𝟐√𝒙𝟐 + 𝒂𝟐 +

𝒂𝟐

𝟐 𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 + 𝒂𝟐| + 𝒄

8. ∫√𝒙𝟐 − 𝒂𝟐 dx = 𝒙

𝟐√𝒙𝟐 − 𝒂𝟐 −

𝒂𝟐

𝟐 𝒍𝒐𝒈 | 𝒙 + √𝒙𝟐 − 𝒂𝟐| + 𝒄

9. ∫√𝒂𝟐 − 𝒙𝟐 dx = 𝒙

𝟐√𝒂𝟐 − 𝒙𝟐 +

𝒂𝟐

𝟐𝐬𝐢𝐧−𝟏 (

𝒂) + 𝒄

INTEGRATION BY PARTS

1. ∫𝒖. 𝒗 𝒅𝒙 = 𝒖. ∫ 𝒗 𝒅𝒙 − ∫( 𝒅𝒖

𝒅𝒙∫𝒗 𝒅𝒙) 𝒅𝒙

The integral of product of two functions = (first function) x integral of the second

function – integral of [(differential coefficient of the first function ) × (integral of the

second function)]

We can choose first and second function according to I L A T E where I → inverse

trigonometric function → 𝒍𝒐𝒈𝒂𝒓𝒊𝒕𝒉𝒎𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏, 𝐀 → 𝒂𝒍𝒈𝒆𝒃𝒓𝒂𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝐓 →

𝒕𝒓𝒊𝒈𝒐𝒏𝒐𝒎𝒆𝒕𝒓𝒊𝒄 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 𝐄 → 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒊𝒂𝒍 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏

2. cxfedxxfxfe xx )()()( 1 .

Working Rule for different types of integrals

1. Integration of trigonometric function

Working Rule

(a) Express the given integrand as the algebraic sum of the functions of the following

(i) Sin k𝜶 , (ii) cos k𝜶 ,(iii) tan k𝜶 , (iv) cot k𝜶 , (v) sec k𝜶 ,(vi) cosec k𝜶 ,(vii) sec2

k𝜶 ,

(viii) cosec2 k𝜶 , (ix) sec k𝜶 tan k𝜶 (x) cosec k𝜶 cot k𝜶

For this use the following formulae whichever applicable

(i) 𝒔𝒊𝒏𝟐 𝒙 = 𝟏−𝒄𝒐𝒔 𝟐𝒙

𝟐 (ii) 𝒄𝒐𝒔𝟐 𝒙 =

𝟏+𝒄𝒐𝒔 𝟐𝒙

(iii) 𝒔𝒊𝒏𝟑𝒙 = 𝟑 𝒔𝒊𝒏 𝒙−𝒔𝒊𝒏 𝟑𝒙

𝟒 (iv)𝒄𝒐𝒔𝟑𝒙 =

𝟑 𝒄𝒐𝒔 𝒙+𝒄𝒐𝒔 𝟑𝒙

(v) tan2 x = sec2 x – 1 (vi) cot2 x = cosec2x – 1

(Vii) 2sin x sin y = cos (x – y ) – cos ( x + y)

(vii) 2 cos x cos y = cos (x + y) + cos (x – y )

(ix) 2 sin x cos y = sin (x + y ) + sin (x – y )

(x) 2cos x sin y = sin (x + y ) – sin ( x – y )

2. Integration by substitution

(a) Consider I= dxxf )(

Put x=g(t) so that ).(1 tgdt

We write dx=g1(t). Thus I= dttgtgfdxxf )())(()( 1

(b) When the integrand is the product of two functions and one of them is a function g

(x) and the other is k g’ (x), where k is a constant then

Put g (x) = t

3. Integration of the types ∫𝒅𝒙

𝒂𝒙𝟐+𝒃𝒙+𝒄 ,∫

𝒅𝒙

√𝒂𝒙𝟐+𝒃𝒙+𝒄, and ∫(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄)𝒅𝒙

In this three forms change 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 in the form A2 + X2 , X2 – A2, or A2 – X2

Where X is of the form x +k and A is a constant ( by completing square method)

Then integral can be find by using any of the special integral formulae

4. Integration of the types ∫𝒑𝒙+𝒒

𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 ,∫

𝒑𝒙+𝒒

√𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 and ∫(𝒑𝒙 +

𝒒)√𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 dx

In this three forms split the linear px +q = 𝝀 𝒅

𝒅𝒙(𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄 ) + 𝝁

Then divide the integral into two integrals

The first integral can be find out by method of substitution and the second integral

by completing square method as explained in 3

I,e., to evaluate ∫𝒑𝒙+𝒒

√𝒂𝒙𝟐+𝒃𝒙+𝒄 𝒅𝒙 = ∫

𝝀 (𝟐𝒂𝒙+𝒃 )+ 𝝁

𝒂𝒙𝟐+ 𝒃𝒙+𝒄 dx

= 𝝀∫𝟐𝒂𝒙+𝒃

𝒂𝒙𝟐+ 𝒃𝒙+𝒄 𝒅𝒙 + 𝝁∫

𝒅𝒙

𝒂𝒙𝟐+ 𝒃𝒙+𝒄

↓ ↓

Find by substitution method + by completing square

method

5. Integration of rational functions

In the case of rational function, if the degree of the numerator is equal or greater

than degree of the denominator , then first divide the numerator by denominator and

write it as 𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓

𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 = 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +

𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓

𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓 , then integrate

6. Integration by partial fractions

Integration by partial fraction is applicable for rational functions. There first we must

check thatdegree of the numerator is less than degree of the denominator, if not, divide

the numerator by denominator and write as 𝑵𝒖𝒎𝒆𝒓𝒂𝒕𝒐𝒓

𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒐𝒓 = 𝑸𝒖𝒐𝒕𝒊𝒆𝒏𝒕 +

𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓

𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓

and proceed for partial fraction of 𝑹𝒆𝒎𝒂𝒊𝒏𝒅𝒆𝒓

𝑫𝒆𝒏𝒐𝒎𝒊𝒏𝒂𝒕𝒊𝒐𝒓

Form of the rational functions Form of the rational functions

1 𝒑𝒙 + 𝒒

(𝒙 − 𝒂)(𝒙 − 𝒃 )

𝒙 − 𝒂+

𝒙 − 𝒃

2 𝒑𝒙 + 𝒒

(𝒙 − 𝒂)𝟐

𝒙 − 𝒂+

(𝒙 − 𝒂)𝟐

3. 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓

(𝒙 − 𝒂 )(𝒙 − 𝒃 )𝒙 − 𝒄 )

𝒙 − 𝒂+

𝒙 − 𝒃+

𝒙 − 𝒄

4 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓

(𝒙 − 𝒂 )𝟐(𝒙 − 𝒃 )

𝒙 − 𝒂+

(𝒙 − 𝒂 )𝟐+

𝒙 − 𝒃

5 𝒑𝒙𝟐 + 𝒒𝒙 + 𝒓

(𝒙 − 𝒂 )(𝒙𝟐 + 𝒃𝒙 + 𝒄 )

Where x2 + bx + c cannot be

factorized further

𝒙 − 𝒂+

𝑩𝒙 + 𝑪

𝒙𝟐 + 𝒃𝒙 + 𝒄

DEFINITE INTEGRATION

Working Rule for different types of definite integrals

1. Problems in which integral can be found by direct use of standard formula or by

transformation method

Working Rule

(i). Find the indefinite integral without constant c

(ii). Then put the upper limit b in the place of x and lower limit a in the place of x and

subtract the second value from the first. This will be the required definite integral.

2. Problems in which definite integral can be found by substitution method

Working Rule

When definite integral is to be found by substitution then change the lower and upper

limits of integration. If substitution is z = φ(x) and lower limit integration is a and

upper limit is b Then new lower and upper limits will be φ(a) and φ(b) respectively.

Properties of Definite integrals

1. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃

𝒂∫ 𝒇(𝒕)𝒅𝒕

𝟐. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃

𝒂∫ 𝒇(𝒙)𝒅𝒙

𝒃. In particular, ∫ 𝒇(𝒙)𝒅𝒙 =

𝒂 0

3. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃

𝒂 + ∫ 𝒇(𝒙)𝒅𝒙

𝒄, a < c < b

4. ∫ 𝒇(𝒙)𝒅𝒙 =𝒃

𝒂∫ 𝒇(𝒂 + 𝒃 − 𝒙)𝒅𝒙

5. ∫ 𝒇(𝒙)𝒅𝒙 =𝒂

𝟎∫ 𝒇(𝒂 − 𝒙)𝒅𝒙

6. ∫ 𝒇(𝒙)𝒅𝒙 =𝟐𝒂

𝟎∫ 𝒇(𝒙)𝒅𝒙

𝟎 + ∫ 𝒇(𝟐𝒂 − 𝒙)𝒅𝒙

7. ∫ 𝒇(𝒙)𝒅𝒙 =𝟐𝒂

𝟎𝟐∫ 𝒇(𝒙)𝒅𝒙

𝟎 , if 𝒇(𝟐𝒂 − 𝒙) = 𝒇(𝒙) and

=0 , if 𝒇(𝟐𝒂 − 𝒙) = −𝒇(𝒙)

8. (i) ∫ 𝒇(𝒙)𝒅𝒙 =𝒂

−𝒂𝟐∫ 𝒇(𝒙)𝒅𝒙

𝟎, if 𝒇 is an even function, i.e., if 𝒇(-x) = 𝒇(𝒙)

(ii) ∫ 𝒇(𝒙)𝒅𝒙 =𝒂

−𝒂𝟎, if 𝒇 is an odd function, i.e., if 𝒇(−x) = −𝒇(𝒙)

Problem based on property

∫ 𝒇(𝒙)𝒅𝒙 =𝒃

𝒂 + ∫ 𝒇(𝒙)𝒅𝒙

𝒄, a < c < b

Working Rule

This property should be used if the integrand is different in different parts of the

interval [a,b] in which function is to be integrand. This property should also be used

when the integrand (function which is to be integrated) is under modulus sign or is

discontinuous at some points in interval [a,b]. In case integrand contains modulus then

equate the functions whose modulus occur to zero and from this find those values of x

which lie between lower and upper limits of definite integration and then use the

property.

∫ 𝒇(𝒙)𝒅𝒙 =𝒂

𝟎∫ 𝒇(𝒂 − 𝒙)𝒅𝒙

Working Rule

Let I = ∫ 𝒇(𝒙)𝒅𝒙𝒂

Then I = ∫ 𝒇(𝒂 − 𝒙)𝒅𝒙𝒂

(1) + (2) => 2I = ∫ 𝒇(𝒙)𝒅𝒙𝒂

𝟎 + ∫ 𝒇(𝒂 − 𝒙)𝒅𝒙

I = 𝟏

𝟐∫ {𝒇(𝒙) + 𝒇(𝒂 − 𝒙)}𝒅𝒙

This property should be used when 𝒇(𝒙) + 𝒇(𝒂 − 𝒙) becomes an integral function of x.

∫ 𝒇(𝒙)𝒅𝒙 =𝒂

−𝒂 0, if 𝒇(𝒙) is an odd function and ∫ 𝒇(𝒙)𝒅𝒙 =

−𝒂 2∫ 𝒇(𝒙)𝒅𝒙

𝟎, if 𝒇(𝒙) is an

even function.

Working Rule

This property should be used only when limits are equal and opposite and the

function which is to be integrated is either odd/ even.

PROBLEM BAESD ON LIMIT OF SUM

Working rule

∫ 𝒇 (𝒙)𝒅𝒙 𝒃

𝒂 = 𝐥𝐢𝐦

𝒉→𝟎𝒉 { 𝒇 (𝒂) + 𝒇 (𝒂 + 𝒉 ) + … .+𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )}

Where nh = b – a

The following results are used for evaluating questions based on limit of sum.

(i) 1 + 2 + 3 + ….. + (n-1) =

(ii) 𝟏𝟐 + 𝟐𝟐 + 𝟑𝟐 + …. + (𝒏 − 𝟏)𝟐 = ∑(𝒏−𝟏)𝒏(𝟐𝒏−𝟏)

(iii) 𝟏𝟑 + 𝟐𝟑 + 𝟑𝟑 + ….. + (𝒏 − 𝟏)𝟑 = [(𝒏−𝟏)𝒏

𝟐]𝟐

(iv) a + ar + …… + 𝒂𝒓𝒏−𝟏 = 𝒂[𝒓𝒏− 𝟏]

𝒓−𝟏 (r≠1)

IMPORTANT SOLVED PROBLEMS

Evaluate the following integrals

1. ∫( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐

𝑿 𝒅𝒙

Solution

put 1+log x = t 𝟏

𝒙 𝒅𝒙 = 𝒅𝒕

∫( 𝟏+𝒍𝒐𝒈 𝑿 )𝟐

𝑿 𝒅𝒙 = ∫ 𝒕𝟐 𝒅𝒕

= 𝒕𝟑

𝟑+ 𝒄

= (𝟏+𝒍𝒐𝒈𝒙 )𝟑

𝟑 + 𝒄

2. ∫𝒆𝒙

√𝟓−𝟒𝒆𝒙−𝒆𝟐𝒙 𝒅𝒙

Solution

Put 𝒆𝒙 = 𝒕 𝒕𝒉𝒆𝒏 𝒆𝒙 𝒅𝒙 = 𝒅𝒕

∫𝒆𝒙

√𝟓 − 𝟒𝒆𝒙 − 𝒆𝟐𝒙 𝒅𝒙 = ∫

𝒅𝒕

√𝟓 − 𝟒𝒕 − 𝒕𝟐

= ∫𝒅𝒕

√− (𝒕𝟐+ 𝟒𝒕−𝟓 )

= ∫𝒅𝒕

√− ( 𝒕𝟐+ 𝟒𝒕+𝟒−𝟒−𝟓 )

= ∫𝒅𝒕

√− { ( 𝒕+𝟐 )𝟐−𝟗 }

= ∫𝒅𝒕

√𝟑𝟐−( 𝒕+𝟐 )𝟐

= 𝐬𝐢𝐧−𝟏 𝒕+𝟐

𝟑 + C = 𝐬𝐢𝐧−𝟏 (

𝒆𝒙+𝟐

𝟑) + 𝑪

3. ∫√𝒕𝒂𝒏𝒙 dx

Solution

Put tan x = t 2then sec2x dx = 2t dt => dx = 𝟐𝒕 𝒅𝒕

𝟏+ 𝒕𝟒

∫√𝒕𝒂𝒏𝒙 dx = ∫ 𝒕 𝟐𝒕 𝒅𝒕

𝟏+ 𝒕𝟒 = ∫

𝟐𝒕𝟐

𝟏+ 𝒕𝟒 𝒅𝒕

= ∫𝟐

𝒕𝟐+ 𝒕𝟐

𝒅𝒕 (by dividing nr and dr by t2 )

= ∫(𝟏+

𝒕𝟐) + (𝟏−

𝒕𝟐)

𝒕𝟐+ 𝟏

𝒕𝟐

𝒅𝒕

= ∫𝟏+

𝒕𝟐

𝒕𝟐+ 𝟏

𝒕𝟐

𝒅𝒕 + ∫𝟏 −

𝒕𝟐

𝒕𝟐+ 𝟏

𝒕𝟐

𝒅𝒕

= ∫𝟏+

𝒕𝟐

(𝒕− 𝟏

𝒕)𝟐+ 𝟐

𝒅𝒕 + ∫𝟏−

𝒕𝟐

(𝒕+ 𝟏

𝒕)𝟐− 𝟐

𝒅𝒕

= ∫𝒅𝒖

𝒖𝟐+ 𝟐 + ∫

𝒅𝒗

𝒗𝟐− 𝟐 ( 1st integral put 𝒕 −

𝒕 = u

then (𝟏 + 𝟏

𝒕𝟐 ) 𝒅𝒕 = 𝒅𝒖

2nd integral put𝒕 +𝟏

𝒕= 𝒗 𝒕𝒉𝒆𝒏 ( 𝟏 −

𝒕𝟐)𝒅𝒕 = 𝒅𝒗

√𝟐𝐭𝐚𝐧−𝟏 (

√𝟐) +

𝟐√𝟐𝒍𝒐𝒈 |

𝒗− √𝟐

𝒗+√𝟐| + 𝑪

= 𝟏

√𝟐𝐭𝐚𝐧−𝟏 (

𝒕−𝟏

√𝟐) +

𝟐√𝟐𝒍𝒐𝒈 |

𝒕+𝟏

𝒕− √𝟐

𝒕+𝟏

𝒕+√𝟐

| + 𝑪

= 𝟏

√𝟐𝐭𝐚𝐧−𝟏 (

𝒕𝟐− 𝟏

√𝟐 𝒕) +

𝟐√𝟐𝒍𝒐𝒈 |

𝒕𝟐+ 𝟏− √𝟐 𝒕

𝒕𝟐+ 𝟏+ √𝟐 𝒕| + 𝑪

= 𝟏

√𝟐𝐭𝐚𝐧−𝟏 (

𝒕𝒂𝒏𝒙− 𝟏

√𝟐 𝒕𝒂𝒏𝒙) +

𝟐√𝟐𝒍𝒐𝒈 |

𝒕𝒂𝒏𝒙+ 𝟏− √𝟐𝒕𝒂𝒏𝒙

𝒕𝒂𝒏𝒙+ 𝟏− √𝟐𝒕𝒂𝒏𝒙| + 𝑪

4. ∫𝟓𝒙+𝟑

√𝒙𝟐+ 𝟒𝒙+ 𝟏𝟎 𝒅𝒙

Solution

5x + 3 = A (2x + 4 ) + B = > A = 𝟓

𝟐 𝒂𝒏𝒅 𝑩 = −𝟕

∫𝟓𝒙 + 𝟑

√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 𝒅𝒙 = ∫

𝟐(𝟐𝒙 + 𝟒) − 𝟕

√𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 𝒅𝒙

= ∫𝟓

𝟐(𝟐𝒙+𝟒)

√𝒙𝟐+ 𝟒𝒙+𝟏𝟎 𝒅𝒙 + ∫

√𝒙𝟐+ 𝟒𝒙+𝟏𝟎𝒅𝒙

= 𝟓

𝟐∫

(𝟐𝒙+𝟒)

√𝒙𝟐+ 𝟒𝒙+𝟏𝟎 𝒅𝒙 + 𝟕 ∫

√𝒙𝟐+ 𝟒𝒙+𝟏𝟎𝒅𝒙

= 𝟓

𝟐∫

𝒅𝒕

√𝒕 + 𝟕 ∫

√𝒙𝟐+ 𝟒𝒙+𝟒−𝟒+𝟏𝟎𝒅𝒙

= 𝟓

𝟐× 𝟐√𝒕 + 𝟕 ∫

√(𝒙+𝟐 )𝟐+𝟔 dx

= 5 √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎 + 𝟕 𝒍𝒐𝒈 | 𝒙 + 𝟐 + √𝒙𝟐 + 𝟒𝒙 + 𝟏𝟎| + 𝑪

5. ∫𝒙 𝒕𝒂𝒏 𝒙

𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙 𝒅𝒙

Solution

I = ∫𝒙 𝒕𝒂𝒏 𝒙

𝟎 -------------------(1)

Again I = ∫(𝝅−𝒙) 𝒕𝒂𝒏(𝝅−𝒙)

𝒔𝒆𝒄 (𝝅−𝒙)+𝒕𝒂𝒏 (𝝅−𝒙) 𝒅𝒙

𝟎 Using the property ∫ 𝒇 (𝒙)𝒅𝒙 = ∫ 𝒇(𝒂 − 𝒙 )𝒅𝒙

I = ∫(𝝅−𝒙) 𝒕𝒂𝒏 𝒙

𝟎 -------------------- (2)

Adding (1) and (2) we get

2 I = ∫𝒙 𝒕𝒂𝒏 𝒙

𝟎+ ∫

(𝝅−𝒙) 𝒕𝒂𝒏 𝒙

= 𝝅∫𝒕𝒂𝒏𝒙

𝒔𝒆𝒄 𝒙+𝒕𝒂𝒏 𝒙𝒅𝒙

𝟎 = 𝝅∫

𝒕𝒂𝒏𝒙 (𝒔𝒆𝒄 𝒙−𝒕𝒂𝒏 𝒙 )

𝒔𝒆𝒄𝟐𝒙− 𝒕𝒂𝒏𝟐𝒙

𝟎 𝒅𝒙

= 𝝅∫ (𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏𝟐𝝅

𝟎𝒙) dx = 𝝅 ∫ (𝒕𝒂𝒏 𝒙 𝒔𝒆𝒄 𝒙 − 𝒔𝒆𝒄𝟐𝒙 + 𝟏)𝒅𝒙

= 𝝅⌈𝒔𝒆𝒄 𝒙 − 𝒕𝒂𝒏 𝒙 + 𝒙⌉𝝅𝟎

I = 𝝅(𝝅

𝟐− 𝟏)

6. Evaluate ∫ (𝒙𝟐𝟒

𝟏− 𝒙) 𝒅𝒙 using limit of sum

Solution

Comparing the given integral with ∫ 𝒇 (𝒙)𝒅𝒙 𝒃

a = 1, b = 4 f (x ) = x2 – x

∴ 𝒏𝒉 = 𝟒 − 𝟏 = 𝟑 and

f ( a+ (n-1)h )= f ( 1 + (n – 1 ) h )

= ( 1 + (n – 1 ) h ) 2 - [1 + (n – 1 ) h]

= 1 + 2 (n-1 )h + (n – 1 ) 2 h2 -1 – (n – 1 ) h

= (n – 1 ) h + (n – 1 ) 2 h2

∫ 𝒇 (𝒙)𝒅𝒙 𝒃

𝒂 = 𝐥𝐢𝐦

𝒉 →𝟎 𝒉 ∑ 𝒇 ( 𝒂 + (𝒏 − 𝟏 )𝒉 )𝒏−𝟏

∫ (𝒙𝟐𝟒

𝟏− 𝒙) 𝒅𝒙 = 𝐥𝐢𝐦

𝒉→𝟎𝒉{∑ (𝒏 − 𝟏 ) 𝒉 + (𝒏 − 𝟏)𝟐𝒏−𝟏

𝟎 𝒉𝟐}

= 𝐥𝐢𝐦𝒉→𝟎

𝒉{∑(𝒏 − 𝟏 )𝒉 + ∑(𝒏 − 𝟏)𝟐𝒉𝟐}

𝒉{𝒉∑(𝒏 − 𝟏 ) + 𝒉𝟐 ∑(𝒏 − 𝟏)𝟐}

𝒉𝟐 𝒏 ( 𝒏−𝟏 )

𝟐 + 𝒉𝟑 𝒏 ( 𝒏−𝟏 )( 𝟐𝒏−𝟏 )

= 𝐥𝐢𝐦𝒉 →𝟎

𝒏𝒉 ( 𝒏𝒉−𝒉 )

𝒏𝒉 ( 𝒏𝒉−𝒉 )(𝟐𝒏𝒉−𝒉 )

𝟑 ( 𝟑−𝒉 )

𝟑 ( 𝟑−𝒉 )( 𝟔−𝒉 )

= 𝟗

𝟐 +

𝟓𝟒

= 𝟐𝟕

PRACTICE PROBLEMS

LEVEL I

1. ∫(𝟐𝒙 − 𝟑 𝒄𝒐𝒔𝒙 + 𝒆𝒙) 𝒅𝒙

2. ∫𝒙𝟒+𝟓𝒙𝟐+ 𝟑

√𝒙 𝒅𝒙

3. ∫ 𝒔𝒊𝒏𝟑 𝒙 𝒅𝒙

4. ∫ 𝒔𝒊𝒏 𝟑𝒙 𝒄𝒐𝒔 𝟓𝒙 𝒅𝒙

5. ∫ 𝒔𝒊𝒏𝟑𝒙𝒄𝒐𝒔𝟓𝒙 𝒅𝒙

6. ∫𝒅𝒙

𝒙𝟐+ 𝟒𝒙+𝟏

7. ∫𝒅𝒙

√𝒙𝟐+ 𝟓𝒙+𝟖

8. ∫ 𝒕𝒂𝒏𝟐𝒙𝝅

𝟒𝟎

𝒅𝒙

9. ∫𝒙𝟐𝒆𝒙 𝒅𝒙

10. ∫𝒙

𝒙𝟐+ 𝟏

𝟎 𝒅𝒙

LEVEL II

1. ∫𝒔𝒊𝒏 𝒙

𝒔𝒊𝒏 (𝒙−𝒂 ) 𝒅𝒙

2. ∫𝟐

( 𝟏−𝒙 )( 𝟏+ 𝒙𝟐) 𝒅𝒙

3. ∫𝒙+𝟐

√( 𝒙−𝟓 )(𝒙−𝟑 ) 𝒅𝒙

4. ∫ 𝒆𝒙 (𝒔𝒊𝒏 𝟒𝒙−𝟒

𝟏−𝒄𝒐𝒔 𝟒𝒙) 𝒅𝒙

5. ∫𝒙𝟐 𝐭𝐚𝐧−𝟏 𝒙 𝒅𝒙

6. ∫𝒅𝒙

𝒄𝒐𝒔𝟐𝒙+𝒔𝒊𝒏 𝟐𝒙

7. ∫𝟐𝒙

( 𝟏+ 𝒙𝟐)( 𝟑+ 𝒙𝟐) 𝒅𝒙

9. ∫ 𝐭𝐚𝐧−𝟏 √𝟏−𝒙

𝟏+𝒙 𝒅𝒙

10. ∫𝒙𝒕𝒂𝒏 𝒙

𝒔𝒆𝒄 𝒙 𝒄𝒐𝒔𝒆𝒄 𝒙

𝟎 𝒅𝒙

11. ∫𝒙

𝟏+𝒔𝒊𝒏𝒙

𝟎 𝒅𝒙

12. ∫ | 𝒙𝟑 − 𝒙 |𝟑

−𝟏 𝒅𝒙

13. ∫√𝒙

√𝒙+ √𝟓−𝒙

𝟏 𝒅𝒙

14. ∫ 𝒍𝒐𝒈 ( 𝟏 + 𝒕𝒂𝒏 𝒙 )𝒅𝒙𝝅

𝟒𝟎

15. ∫ (𝒙𝟐 + 𝟐) 𝒅𝒙 𝟐

𝟎 as a limit of sum

dxCosxSinx

CosxSinx

LEVEL III

1. ∫𝒔𝒊𝒏 𝒙

𝒔𝒊𝒏𝟒𝒙 𝒅𝒙

2. ∫(√𝒕𝒂𝒏 𝒙 + √𝒄𝒐𝒕 𝒙 ) 𝒅𝒙

3. ∫𝒔𝒊𝒏 𝒙 𝒄𝒐𝒔 𝒙

𝒔𝒊𝒏𝟒𝒙+ 𝒄𝒐𝒔𝟒𝒙 𝒅𝒙

4. ∫√𝟏− √𝒙

𝟏+√𝒙 𝒅𝒙

5. ∫ 𝒔𝒊𝒏 (𝒍𝒐𝒈 𝒙 )𝒅𝒙

6. ∫ 𝒍𝒐𝒈 ( 𝒙 + √𝒙𝟐 + 𝒂𝟐) 𝒅𝒙

7 ∫ 𝒙 𝒍𝒐𝒈 (𝒔𝒊𝒏 𝒙 )𝒅𝒙𝝅

8. ∫ (𝟐𝒙𝟐 + 𝒙 + 𝟕 ) 𝒅𝒙𝟐

𝟏 as a limit of sum

)1)(1( 2

1loglog

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