# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.30 | Set 3

• Last Updated : 30 Apr, 2021

### Question 41. ∫ x²/(x²+1)(3x²+4) dx

Solution:

Let x²/(x²+1)(3x²+4) =(Ax+B)/(x²+1) +(Cx+D)/(3x²+4)

x²=(Ax+8)(3x²+4)+(Cx+D)(x²+1)

=(3A+C)x3+(3B+D)x²+(4A+C)x+4B+D

Equating similar terms, we get,

3A+C=0,

3B+D=1,

4A+C=0,

4B+D=0

Solving, we get,

A=0,

B=-1,

C=0,

D=4

Thus,

I =∫ (-dx)/(x²+1) +∫ 4dx/((3x²+4))

=-tan-1⁡x+4/3 ∫ dx/(x²+(2/√3)²)

=-tan-1⁡x+(4/3)*(√3/2)tan-1((√3 x)/2)+c

I =2/√3 tan-1((√3 x)/2)-tan-1⁡x+c

### Question 42. ∫(3x+5)/(x3-x²-x+1) dx

Solution:

∫ (3x+5)/(x3-x²-x+1) dx

Let (3x+5)/((x-1)² (x+1))=A/(x-1)+B/((x-1)²)+C/(x+1)

3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)²

Put x=1

B=4

Put x=-1

C=1/2

Put x=0

A=-1/2

Therefore

∫ (3x+5)/((x-1)² (x+1)) dx

=-1/2 ∫ dx/(x-1)+4∫ dx/((x-1)²)+1/2 ∫ dx/(x+1)

=-1/2 ln⁡|(x-1)|-4/((x-1))+1/2 ln⁡|(x+1)|+c

=1/2 ln⁡|(x+1)/(x-1)|-4/((x-1))+c

### Question 43. ∫ (x3-1)/(x3+x) dx

Solution:

Let I =∫ (x3-1)/(x3+x) dx

=∫(1-(x+1)/(x3+x))dx

=∫ dx-∫ (x+1)/(x3+x) dx

Let (x+1)/x(x²+1) =A/x+(Bx+C)/(x²+1)

x+1 =A(x²+1)+(Bx+C)x

=(A+8)x²+(B+C)x+A

Equating similar terms, we get,

A+B=0,

C=1,

A=1

Solving, we get,

A=1,

B=-1,

C=1

Thus,

I=-∫ dx/x-∫ (-x+1)/(x²+1) dx+∫ dx

I =x-log⁡|x|+1/2 log⁡|x²+1|-tan-1⁡x+c

I=x-log⁡|x|+1/2 log⁡|x²+1|-tan-1⁡x+c

### Question 44. ∫ (x²+x+1)/((x+1)² (x+2)) dx

Solution:

∫ (x²+x+1)/((x+1)² (x+2)) dx

Let (x²+x+1)/((x+1)² (x+2))=A/(x+1)+B/((x+1)²)+C/(x+2)

x²+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)²

Putting x=-1

B=1

Putting x=-2

C=3

Putting x=0

A=-2,

Therefore,

∫ (x²+x+1)/((x+1)² (x+2)) dx

=-2∫ dx/(x+1)+∫ dx/((x+1)²)+3∫ dx/(x+2)

=-2ln⁡|x+1|-1/(x+1)+3ln⁡|x+2|+c

### Question 45. ∫ 1/x(x4+1) dx

Solution:

Let 1/x(x4+1) =A/x+(Bx3+Cx2+Dx+E)/(x4+1)

1=A(x4+1)+(Bx3+Cx²+Dx+E)x

=(A+B)x4+Cx3+Dx²+Ex+A

Equating similar terms, we get,

A+B=0,

C=0,

D=0,

E=0,

A=1,

B=-1

Thus,

I =∫ dx/x+∫ -(x3 dx)/(x4+1)

=log⁡|x|-1/4 log⁡|x4+1|+c

I =1/4 log⁡|x4/(x4+1)|+c

### Question 46. ∫1/(x(x3+8)) dx

Solution:

I=∫1/(x(x3+8)) dx

Arranging the above equation,

I=∫ x²/(x3 (x3+8)) dx

=1/3∫(3x²)/(x3 (x3+8)) dx

Now substituting x3=t, we have, 3x3 dx=dt

I=1/3∫dt/(t(t+8))

Integrand by partial fractions. Thus,

1/(t(t+8))=A/t+B/(t+8)

1/(t(t+8))=(A(t+8)+Bt)/(t(t+8))

1=A(t+8)+Bt

1=At+8A+Bt

Comparing the coefficients, we have,

A+B=0,

8A=-1/8

Therefore,

I=1/3 ∫ dt/(t(t+8))

=1/3∫ {(1/8)/t – (1/8)/(t+8)} dt

=(1/3)*(1/8) ∫ dt/t – (1/3)*(1/8)∫ dt/(t+8)

=(1/24)* log | t |-(1/24)* log| t+8 |+c

=(1/24) log| x3 |-(1/24) log| x3+8|+c

=(3/24) log|x3|- (1/24)log|x3+8|+c

I=(1/8) log| x3| – (1/24)log|x3+8|+c

### Question 47. ∫ 3/((1-x)(1+x²)) dx

Solution:

Let 3/((1-x)(1+x²))=A/(1-x)+(Bx+C)/(1+x²)

3 =A(1+x²)+(Bx+C)(1-x)

=(A-B)x²+(B-C)x+(A+C)

Equating similar terms, we get,

A-B=0,

B-C=0,

A+C=3

Solving we get,

A=C=3/2 and B=3/2

Thus,

I=3/2 ∫ dx/(1-x)+3/2 ∫ xdx/(1+x²)+3/2 ∫ dx/(1+x²)

=-3/2 log⁡|1-x|+3/2 log⁡|1+x² |+3/2 tan-1⁡x+c

I=3/4 [log⁡|(1+x²)/((1-x)²)|+2tan-1⁡x+c

### Question 48. ∫ (cos⁡x)/((1-sin⁡x)3(2+sin⁡x)) dx

Solution:

Let sin⁡x=t

cos⁡x=dt

∫ (cos⁡x)/((1-sin⁡x)3(2+sin⁡x))=∫ 1/((1-t)3 (2+t)) dt

Let f(t)=1/((1-t)3 (2+t))

Then, suppose

1/((1-t)3 (2+t)) = A/(1-t)+B/((1-t)²)+c/((1-t)3)+D/((2+t))

1=A(1-t)² (2+t)+B(1-t)(2+t)+C(2+t)+D(1-t)3

Put t=1

1=3C

C=1/3

Put t=-2

1=27D

D=1/27

Similarly, we can find that A=(-1)/27 and B=(+1)/9

∫ 1/((1-t)3 (2+t)) dt

=(-1)/27 ∫ 1/(1-t) dt+1/9 ∫ dt/((1-t)²)+1/3 ∫ dt/((1-t)3)+1/27 ∫ dt/(2+t)

=(-1)/27 log⁡|1-t|+1/(9(1-t))+1/(6(1-t)²)+1/27 log⁡|2+t|+c

Putting t=sin⁡x , we get

∫ (cos⁡x)/((1-sin⁡x)3 (2+sin⁡x)) dx

=(-1)/27 log⁡|1-sin⁡x|+1/(9(1-sin⁡x))+1/(6(1-sin⁡x)²)+1/27 log⁡|2+sin⁡x|+c

### Question 49. ∫(2x²+1)/(x² (x²+4)) dx

Solution:

I=∫(2x²+1)/(x² (x²+4)) dx

Now let us separate the fraction (2x²+1)/(x² (x²+4)) through partial fractions.

Substitute x²=t, then (2x²+1)/(x² (x²+4))=(2t+1)/(t(t+4))

(2t+1)/(t(t+4))=A/t+B/(t+4)

(2t+1)/(t(t+4))=(A(t+4)+Bt)/(t(t+4))

2t+1=A(t+4)+Bt

2t+1=At+4A+Bt

Comparing the coefficients, we have, A+B=2 and 4A=1

A=1/4 and B=7/4

(2x²+1)/(x² (x²+4))=1/(4x²)+7/4(x²+4)

Thus ,we have

I= ∫ (2x²+1)/(x²(x²+4)) dx

=1/4 ∫ dx/x² dx +7/4 ∫ dx/(x²+4) dx

=-1/4x +(7/4)*(1/2) tan-1(x/2)+c

I=-1/4x+(7/8)tan-1(x/2)+c

### Question 50. ∫ cosx/(1-sinx)/(2-sinx) dx

Solution:

∫ cosx/(1-sinx)/(2-sinx) dx

Let 1-sinx=t and

-cos x dx = dt

Therefore

-∫ dt/(t(1+t))=-∫ (1/t-1/(t+1))dt

=ln⁡|(t+1)|-ln⁡|t|+c

=ln⁡|(t+1)/t|+c

=ln⁡|(2-sin⁡x)/(1-sin⁡x)|+c

### Question 51. ∫(2x+1)/((x-2)(x-3)) dx

Solution:

Let (2x+1)/((x-2)(x-3))=A/((x-2))+B/(x-3)

2x +1=A(x-3)+B(x-2)

=(A+B)x+(-3A-2B)

Equating similar terms, we get,

A+B=2, and -3A-2B=1

Thus,

I=-5∫ dx/(x-2)+7∫ dx/(x-3)

=-5log⁡|x-2|+7log⁡|x-3|+c

I=log⁡|((x-3)7/((x-2)5)|+c

### Question 52. ∫ 1/((x²+1)(x²+2)) dx

Solution:

Let x²=y

Then 1/((y+1)(y+2))=A/(y+1)+B/(y+2)

1 =A(y+2)+B(y+1)

=(A+B)y+(2A+B)

Equating similar terms, we get,

A+B=0, and 2A+B=1

Solving, we get,

Thus,

I=∫ dx/(x²+1)-∫ dx/(x²+2)

I=tan⁡-1x-1/√2 tan-1⁡x/√2+c

### Question 53. ∫ 1/x(x4-1) dx

Solution:

∫ 1/x(x4-1) dx

Let 1/x(x4-1) =A/x+B/(x+1)+C/(x-1)+D/(x²+1)

1=A(x+1)(x-1)(x²+1)+Bx(x-1)(x²+1)+Cx(x+1)(x²+1)+Dx(x+1)(x-1))

Put x=0

A=-1,

Put x=1

C=1/4

Put x=-1

B=1/4

Put x=2

D=1/4

Therefore

∫ 1/x(x4-1) dx

=-∫ 1/x dx+1/4 ∫ dx/(x+1)+1/4 ∫ dx/(x-1)+1/4 ∫ dx/(x²+1)

=-ln⁡|x|+1/4 ln⁡|(x+1)|+1/4 ln⁡|(x-1)|+1/4 ln⁡|(x²+1)|+c

=1/4 ln⁡|(x4-1)/x4 |+c

### Question 54. ∫ 1/(x4-1) dx

Solution:

∫ 1/(x4-1) dx

Let 1/(x4-1) =A/(x+1)+B/(x-1)+C/(x²+1)

1=A(x-1)(x²+1)+B(x+1)(x²+1)+C(x+1)(x-1)

Put x=1

B=1/

Put x=-1

A=-1/4

Put x=0

C=-1/2

Therefore,

∫ 1/(x4-1) dx

=-1/4 ∫ dx/(x+1)+1/4 ∫ dx/(x-1)-1/2 ∫ dx/(x²+1)

=-1/4 ln⁡|(x+1)|+1/4 ln⁡|(x-1)|-1/2 tan-1⁡x+c

=1/4 ln⁡|(x-1)/(x+1)|-1/2 tan-1x+c

### Question 55. ∫ dx/(cos⁡x(5-4sin⁡x)) dx

Solution:

Let I=∫ dx/(cos⁡x(5-4sin⁡x))

=∫ (cos⁡xdx)/(cos²⁡x(5-4sin⁡x))

=∫ (cos⁡xdx)/((1-sin²⁡x)(5-4sin⁡x))

Let sin⁡x=t

cos ⁡x dx = dt

I=∫ dt/((1-t²)(5-4t))

Now,

Let 1/((1-t²)(5-4t))=A/(1-t)+B/(1+t)+C/(5-4t)

1=A(1+t)(5-4t)+B(1-t)(5-4t)+C(1-t²)

Put t=1

1=2A

A=1/2

Put t=-1

1=18B

B=1/18

Put t=5/4

1=-9C/16

C=-16/9

Thus,

I=1/2∫dt/(1-t)+1/18∫dt/(1+t)-16/9∫dt/(5-4t)

=-1/2 log⁡|1-t|+1/18 log⁡|1+t|+4/9 log⁡|5-4t|+c

Hence,

I=-1/2 log⁡|1-sin⁡x|+1/18 log⁡|1+sin⁡x|+4/9 log⁡|5-4sin⁡x|+c

### Question 56. ∫ 1/(sinx(3+2cos⁡x)) dx

Solution:

Let I =∫ 1/(sinx(3+2cos⁡x)) dx

=∫ (sin⁡xdx)/(sin²x(3+2cos⁡x))

=∫ (sin⁡xdx)/((1-cos²x)(3+2cos⁡x))

Let cos⁡x=t

-sin⁡xdx=dt

I=∫ dt/((t²-1)(3+2t))

Now,

Let 1/((t²-1)(3+2t))=A/(t-1)+B/(t+1)+C/(3+2t)

1=A(t+1)(3+2t)+B(t-1)(3+2t)+C(t²-1)

Put t=1

1=10A

A=1/10

Put t=-1

1=-2B

B=-1/2

Put t=-3/2

1=5/4C

C=4/5

Thus,

I=1/10∫dt/(t-1)-1/2∫dt/(t+1)+5/4∫dt/(3+2t)

=1/10 log⁡|t-1|-1/2 log⁡|t+1|+2/5 log⁡|3+2t|+c

Hence,

I=1/10 log⁡|cos⁡x-1|-1/2 log⁡|cos⁡x+1|+2/5 log⁡|3+2cos⁡x|+c

### Question 57. ∫1/(sin⁡x+sin⁡2x) dx

Solution:

Let I=∫1/(sin⁡x+sin⁡2x) dx

=∫ dx/(sin⁡x+2sin⁡xcos⁡x)

=∫ (sin⁡xdx)/((1-cos²⁡x)+2(1-cos²x)cos⁡x)

Let cos⁡x=t

-sin⁡xdx=dt

I =∫ dt/((t²-1)+2(t²-1)t)

=∫ dt/((t²-1)(1+2t))

Let ∫1/((t²-1)(1+2t))=A/(t-1)+B/(t+1)+C/(1+2t)

1=A(t+1)(1+2t)+B(t-1)(1+2t)+c(t²-1)

Put t=1

1=6A

A=1/6

Put t=-1

1=2B

B=1/2

Put t=-1/2

1=-3/4 C

C=-4/3

Thus,

I=1/6∫dt/(t-1)+1/2∫dt/(t+1)-4/3∫dt/(1+2t)

=1/6 log⁡|t-1|+1/2 log⁡|t+1|-2/3 log⁡|1+2t|+c

Hence,

I=1/6 log|cosx-1| +1/2 log|cosx+1| -2/3 log|1+2cosx| +c

### Question 58. ∫(x+1)/x(1+xex) dx

Solution:

Let I=∫(x+1)/x(1+xex) dx

=∫((x+1)(1+xex-xex))/x(1+xex) dx

=∫((x+1)(1+xex))/x(1+xex) dx – ∫((x+1)(xex))/x(1+xex) dx

=∫((x+1))/x dx-∫(ex (x+1))/(1+xex) dx

=∫((x+1)ex)/(xex) dx-∫(ex (x+1))/(1+xex) dx

=log⁡|xex |-log⁡|1+xex |+c

I=log⁡|(xex)/(1+xex)|+c

### Question 59. ∫ (x²+1)(x²+2)/(x²+3)(x²+4) dx

Solution:

f(x)=(x²+1)(x²+2)/(x²+3)(x²+4)

Now,

((x²+1)(x²+2)/(x²+3)(x²+4)

=(x4+3x2+2)/(x4+7x2+12)

=((x4+7x²+12)-4x²-10)/(x4+7x²+12)

=1-(4x²+10)/(x4+7x²+12)

Now,

(4x²+10)/(x4+7x²+12)

=(4x²+10)/(x²+3)(x²+4)

Let (4x²+10)/(x²+3)(x²+4) =(Ax+B)/(x²+3)+(Cx+D)/(x²+4)

4x²+10=(Ax+B)(x²+4)+(Cx+D)(x²+3)

Let x=0, we get

10=48+3D—————————–(i)

If x=1, we get

14=5(A+B)+4(C+D)=5A+5B+4C+4D——————(ii)

if x=-1, we get

14=5(-A+B)+4(-C+D)=-5A+5B-4C+4D—————-(iii)

Applying (ii) and (iii) we get,

28=10B+8D

1=5B+4D ——————————–(iv)

From (i) we get,

10=4B+3D

Multiplying equation (iv) by 3 and (i) by 4 and subtracting, we get

42-40=15B-16B

2=-B

B=-2

Putting value of B in (i), we get

10=4(-2)+3D

(10+8)/3=D

D=6

Comparing coefficients of x3 in

4x²+10=(Ax+B)(x²+4)+(Cx+4)(x²+3),

we get

0=A+C

Comparing coefficients of x, we get

0=4A+3C

A=C=0

f(x)=1-(-2)/(x²+3)-6/(x²+4)

=1+2/(x²+3)-6/(x²+4)

∫ f(x)dx=∫ 1+2/(x²+3)-6/(x²+4) dx

=x+2/√3 tan-1x/√3-3tan-1⁡x/2+c

### Question 60. ∫ (4×4+3)/ ((x²+2)(x²+3)(x²+4)) dx

Solution:

let x²=y

(4x4+3)/(x²+2)(x²+3)(x²+4)

=(4y²+3)/((y+2)(y+3)(y+4))

Now,

Let (4y²+3)/((y+2)(y+3)(y+4))=A/(y+2)+B/(y+3)+C/(y+4)

4y²+3 =A(y+3)(y+4)+B(y+2)(y+4)+c(y+2)(y+3)

=(A+B+C)y²+(7A+6A+5C)y+12A+8B+6C

Equating similar terms,

A+B+C=4,

7A+6A+5C=0,

12A+8B+6C=3

Solving, we get

A=19/2,

B=-39,

C=67/2

Thus,

I=19/2 ∫ dx/(x²+2)+(-39)∫ dx/(x²+3)+67/2 ∫ dx/(x²+4)

I=19/(2√2) tan-1⁡(x/√2)-39/√3 tan-1⁡(x/√3)+67/4 tan-1⁡(x/2)+c

Hence,

I=19/(2√2) tan-1⁡(x/√2)-39/√3 tan-1⁡(x/√3)+67/4 tan-1⁡(x/2)+c

### Question 61. ∫ x4/((x-1)(x²+1)) dx

Solution:

x4/((x-1)(x²+1))=x4/(x3-x²+x-1)

=(x(x3-x²+x-1)+1(x3-x²+x-1)+1)/((x3-x²+x-1))

=x+1+1/((x-1)(x²+1))

Now,

1/((x-1)(x²+1))=A/(x-1)+(Bx+C)/(x²+1)

1=A(x²+1)+(8x+C)(x-1)

Put x=1

1=2A

A=1/2

Put x=0

1=A-C

C=A-1=-1/2

Put x=-1

1=2A+2B-2C

=2(A-C)+2B

1=2+2B

2B=-1

B=-1/2

∫ x4/((x-1)(x²+1)) dx

=∫ xdx+∫ 1dx+1/2 ∫ 1/(x-1) dx-1/2 ∫ (x+1)/(x²+1) dx

=x²/2+x+1/2 log⁡|x-1|-1/4 log|⁡(x²+1)|-1/2 tan-1⁡x+c

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