# RD Sharma Class 12 Ex 19.30 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.30 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.30 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.30 Solutions Chapter 19 Indefinite Integrals

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

Solution:

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

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

Put x=2

5=3B⇒ B=5/3

Put x=-1

-1=-3A ⇒ A=1/3

So,

∫ (2x+1)/((x+1)(x-2)) dx

=1/3 ∫ dx/(x+1)+5/3 ∫ dx/(x-2)

=1/3 log⁡|x+1|+5/3 log⁡|x-2|+c

Thus,

I=1/3 log⁡|x+1|+5/3 log⁡|x-2|+c

### Question 2. ∫ 1/(x(x-2)(x-4)) dx

Solution:

Let ∫ 1/(x(x-2)(x-4)) dx=A/x+B/(x-2)+C/(x-4)

1=A(x-2)(x-4)+B(x)(x-4)+C×(x-2)

Put x=0

1=8A ⇒ A=1/8

Put x=2

1=-4B ⇒ B=-1/4

Put x=4

1=8C ⇒ C=1/8

So,

∫ 1/(x(x-2)(x-4)) dx=1/8 ∫ dx/x+(-1/4) ∫ dx/(x-2)+1/8 ∫ dx/(x-4)

=1/8 log⁡|x|-1/4 log⁡|x-2|+1/8 log⁡|x-4|+c

=1/8 log⁡|(x(x-4))/((x-2)2 )|+c

### Question 3. ∫(x2+x-1)/(x2+x-6) dx

Solution:

Let I=∫(x2+x-1)/(x2+x-6) dx

=∫ [1+5/(x2+x-6) ]dx

I=∫ dx+∫ 5dx/((x+3)(x-2))

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

5=A(x-2)+8(x+3)

Put x=2

5=5B ⇒ B=1

Put x=-3

5=-5A ⇒ A=-1

I=∫ dx+∫ (-dx)/(x+3)+∫ dx/(x-2)@

=x-log⁡|x+3|+log⁡|x-2|+c)

Hence,

I=x-log⁡|x+3|+log⁡|x-2|+c

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

Solution:

Let I=∫(3+4x-x2)/(x+2)(x-1) dx

=∫[-1d+(5x+1)/(x+2)(x-1) ] dx

I=-∫dx+∫(5x+1)/(x+2)(x-1) dx

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

5x+1=A(x-1)+B(x+2)

Put x=1

6=3B ⇒ B=2

Put x=-2

-9=-3A ⇒ A=3

So, I=-∫dx+3∫dx/(x+2)+2 ∫dx/(x-1)

I=-x+3log⁡|x+2|+2log⁡|x-1|+c

### Question 5. ∫(x2+1)/(x2-1) dx

Solution:

Let I =∫(x2+1)/(x2-1) dx

=∫[1+2/(x2-1) ]dx

=∫dx+∫2dx/(x+1)(x-1)

=∫dx+∫(-1)/(x+1)+1/(x-1) dx

=x-log⁡|x+1|+log⁡|x-1|+c

I=x +log⁡|(x-1)/(x+1)|+c

### Question 6. ∫x2/(x-1)(x-2)(x-3) dx

Solution:

Let I=∫x2/(x-1)(x-2)(x-3)=A/(x-1)+B/(x-2)+C/(x-3)

x2=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)

Put x=1

1=2A ⇒ A=1/2

Put x=2

4=-B ⇒ B=-4

Put x=3

9=2C ⇒ C=9/2

Thus, I=∫x2/(x-1)(x-2)(x-3) dx=1/2∫dx/(x-1)-4j dx/(x-2)+9/2∫dx/(x-3)

=1/2 log⁡|x-1|-4log⁡|x-2|+9/2 log⁡|x-3|+c

Hence,

I=1/2 log⁡|x-1|-4log⁡|x-2|+9/2 log⁡|x-3|+c

### Question 7. ∫ 5x/(x+1)(x2-4) dx

Solution:

5x/(x+1)(x2-4) =5x/(x+1)(x+2)(x-2)

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

5x=A(x+2)(x-2)+B(x+1)(x-2)+C(x+1)(x+2) ————————–(i)

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

A=5/3 , B=-5/2 , and C=5/6

5x/((x+1)(x+2)(x-2))=5/(3(x+1))-5/(2(x+2))+5/(6(x-2))

∫ 5x/(x+1)(x2-4) dx

=5/3 ∫ 1/(x+1) dx-5/2 ∫ 1/(x+2) dx+5/6 ∫ 1/(x-2) dx

=5/3 log⁡|x+1|-5/2 log⁡|x+2|+5/6 log⁡|x-2|+c

### Question 8. ∫(x2+1)/x(x2-1) dx

Solution:

Let I=∫(x2+1)/x(x2-1) dx=∫(x2+1)/(x(x+1)(x-1)) dx

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

x2+1=A(x+1)(x-1)+B⋅x(x-1)+Cx(x+1)

Put x=0

1=-A ⇒ A=-1

Put x=-1

2=2B ⇒ B=1

Put x=1

2=2C ⇒ C=1

Thus, I=-∫dx/x+∫dx/(x+1)+∫dx/(x-1)

=-log⁡|x|+log⁡|x+1|+log⁡|x-1|+c

I=log⁡|(x2-1)/x|+c

### Question 9. ∫(2x-3)/(x2-1)(2x+3) dx

Solution:

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

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

2x-3=A(x-1)(2x+3)+B(x+1)(2x+3)+C(x2-1)

Put x=-1

-5=-2A

A=5/2

Put x=1

-1=10B ⇒ B=-1/10

Put x=-3/2

-6=5/4 C ⇒ C=-24/5

Thus,

I =5/2 ∫ dx/(x+1)-1/10 ∫ dx/(x-1)-24/5 ∫ dx/(2x+3)

=5/2 log⁡|x+1|-1/10 log⁡|x-1|-24/5(1/2 log⁡|2x+3|)+c

Hence,

I=5/2 log⁡|x+1|-1/10 log⁡|x-1|-12/5 log⁡|2x+3|+c

### Question 10. ∫ x3/(x-1)(x-2)(x-3) dx

Solution:

Let I =∫ x3/(x-1)(x-2)(x-3) dx

=∫ [ 1+(6x2-9x+6)/(x-1)(x-2)(x-3)] dx

Let (6x2-11x+6)/(x-1)(x-2)(x-3)=A/(x-1)+B/(x-2)+C/(x-3)

6x2-11x+6=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)

Put x=1

1=2A ⇒ A=1/2

Put x=2

8=-B ⇒ B=-8

Put x=3

27=2C ⇒ C=27/2

Thus, I=∫dx+1/2∫dx/(x-1)-8∫dx/(x-2)+27/2∫dx/(x-3)

=x+1/2 log⁡|x-1|-8log⁡|x-2|+27/2 log⁡|x-3|+c

Hence,

I=x+1/2 log⁡|x-1|-8log⁡|x-2|+27/2 log⁡|x-3|+c

### Question 11. ∫(sin⁡2x)/(1+sin⁡x)(2+sin⁡x) dx

Solution:

Let ∫(sin⁡2x)/(1+sin⁡x)(2+sin⁡x) dx=A/(1+sin⁡x)+B/(2+sin⁡x)

sin⁡2x=A(2+sin⁡x)+B(1+sin⁡x)

2sin⁡xcos⁡x=(2A+B)+(A+B)sin⁡x

Equating similar terms, we get,

2A+B=0 ⇒ B=-2A and

A+B=2cos⁡x ⇒ -A=2cos⁡x

A=-2cos⁡x and B=+4cos⁡x

Thus,

I=∫-(2cos⁡x)/(1+sin⁡x)dx+∫(4cos⁡x)/(2+sin⁡x)dx

=-2log⁡|1+sin⁡x|+4log⁡|2+sin⁡x|+c

I=log⁡|((2+sin⁡x)4)/((1+sin⁡x)2 )|+c

### Question 12. ∫2x/(x2+1)(x2+3) dx

Solution:

Let ∫2x/(x2+1)(x2+3) dx=(Ax+8)/(x2+1)+(cx+D)/(x2+3)

2x =(Ax+B)(x2+3)+(Cx+D)(x2+1)

=(A+C)x3+(B+D)x2+(3A+C)x+(3B+D))

Equating similar terms, we get,

A+C=0,B+D=0,3A+C=2 and 3B+D=0

A=-C, B=D=0

2A=2 ⇒ A=1 and C=-1

Thus,

I=∫xdx/(x2+1)-∫xdx/(x2+3)

=1/2 log⁡|x2+1|-1/2 log⁡|x2+3|+c

I=1/2 log⁡|(x2+1)/(x2+3)|+c

### Question 13. ∫1/(xlog⁡x(2+log⁡x)) dx

Solution:

Let ∫1/(xlog⁡x(2+log⁡x))=A/(xlog⁡x)+B/(x(2+log⁡x))

1=A(2+log⁡x)+8log⁡x

Put x=1

1=2A ⇒ A=1/2

Put x=10-2

1=-2B ⇒ B=-1/2

I=1/2∫dx/(xlog⁡x)+(-1/2)∫dx/(x(2+log⁡x))

=1/2 log⁡|log⁡x|-1/2 log⁡|2+log⁡x|+c

I=1/2 log⁡|(log⁡x)/(2+log⁡x)|+c

### Question 14. ∫ (ax2+bx+c)/((x-a)(x-b)(x-c)) dx

Solution:

Let (ax2+bx+c)/((x-a)(x-b)(x-c))=A/(x-a)+B/(x-b)+C/(x-c)

ax2+bx+c=A(x-b)(x-c)+B(x-a)(x-c)+c(x-a)(x-b)

Put x=a

a3+ba+c=(a-b)(a-c)A ⇒ A=(a3+ba+c)/((a-b)(a-c))

Put x=b

ab2+b2+c=(b-a)(b-c)B ⇒ B=(ab2+b2+c)/((b-a)(b-c))

Put x=c

ac2+bc+c=(c-a)(c-b)c ⇒ c=(ac2+bc+c)/((c-a)(c-b))

I=(a3+ba+c)/((a-b)(a-c))∫dx/(x-a)+(ab2+b2+c)/((b-a)(b-c))∫dx/(x-b)+(ac2+bc+c)/((c-a)(c-b))∫dx/(x-c)

Hence,

I=(a3+ba+c)/((a-b)(a-c)) log⁡|x-a|+(ab2+b2+c)/((b-a)(b-c)) log⁡|x-b|+(ac2+bc+c)/((c-a)(c-b)) log⁡|x-c|+c

### Question 15. ∫ x/((x2+1)(x-1)) dx

Solution:

Consider the integral

I=∫ x/((x2+1)(x-1)) dx

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

x/(x2+1)(x-1)=A/(x-1)+(Bx+C)/(x2+1)2

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

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

x=Ax2+A+Bx2-Bx+Cx-C

Comparing the co-efficients, we have,

A+B=0,

-B+C=1 and

A-C=0

Solving the equations, we get,

A=1/2 ,

B=-1/2 and C=1/2

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

x/((x2+1)(x-1))=1/2×1/(x-1)-1/2×(x-1)/(x2+1)

x/((x2+1)(x-1))=1/(2(x-1))-x/2(x2+1) +1/2(x2+1)

Thus, we have, I=∫x/((x2+1)(x-1)) dx

=∫[1/(2(x-1))-x/2(x2+1) +1/2(x2+1)]dx

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

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

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

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

### Question 16. ∫ 1/(x-1)(x+1)(x+2) dx

Solution:

Let I=∫1/(x-1)(x+1)(x+2)=A/(x-1)+B/(x+1)+C/(x+2)

1=A(x+1)(x+2)+B(x-1)(x+2)+c(x2-1)

Put x=1

1=6A ⇒ A=1/6

Put x=-1

1=-2B ⇒ B=-1/2

Put x=-2

1=3C ⇒ C=1/3

So,

I=1/6∫dx/(x-1)-1/2∫dx/(x+1)+1/3∫dx/(x+2)

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

### Question 17. ∫x2/(x2+4)(x2+9) dx

Solution:

Consider the integral

I=∫x2/(x2+4)(x2+9) dx

Now let us separate the fraction x2/(x2+4)(x2+9)

through partial fractions.

Substitute x2=t

x2/(x2+4)(x2+9) =t/(t+4)(t+9)

t/(t+4)(t+9) = A/(t+4) + B/(t+9)

t/(t+4)(t+9) = (A(t+9)+B(t+4))/(t+4)(t+9)

t=A(t+9)+B(t+4)

t=At+9A+Bt+4B

Comparing the coefficients, we have,

A+B=1 and 9A+4B=0

A=-4/5 and B =9/5

x2/(x2+4)(x2+9) =-4/(5(t+4))+9/(5(t+9))

x2/(x2+4)(x2+9) =-4/5(x2+4) +9/5(x2+9)

Thus, we have,

I=∫x2/(x2+4)(x2+9) dx

=∫[-4/5(x2+4) +9/5(x2+9) ]dx

=-∫4dx/5(x2+4) +∫9dx/5(x2+9)

=-4/5∫dx/(x2+4) +9/5∫dx/(x2+9)

=-4/5×1/2 tan-1(x/2)+9/5×1/3 tan-1⁡(x/3)+C

=-2/5 tan-1(x/2)+3/5 tan-1(x/3)+C

### Question 18. ∫(5x2-1)/x(x-1)(x+1) dx

Solution:

Let ∫(5x2-1)/x(x-1)(x+1) dx=A/x+B/(x-1)+C/(x+1)

5x2-1=A(x2-1)+B(x+1)x+C(x-1)x

Put x=0

-1=-A ⇒ A=1

Put x=+1

4=2B ⇒ B=2

Put x=-1

4=2C ⇒ C=2

So,

I=∫dx/x+∫2dx/(x-1)+∫2dx/(x+1)

=log⁡|x|+2log⁡|x-1|+2log⁡|x+1|+c

I=log⁡|x(x2-1)2 |+c

### Question 19. ∫(x²+6x-8)/(x3-4x)dx

Solution:

Let I=∫(x²+6x-8) /x(x+2)(x-2) dx

Now,

let (x²+6x-8)/x(x+2)(x-2)=A/(x)+B/(x+2)+C/(x-2)

x²+6x-8=A(x²-4)+B(x-2)+C(x(x+2))

Put x=0

-8=-4A

A=2

Put x=-2

-16=8B

B=-2

Put x=2

8=8C

c=1

Thus,

I=∫2dx/x-∫dx/(x+2)+∫dx/(x-2)

=2log|x|-log|x+2|+log|x-2|+c

I=log|(x²(x-2))/(x+2)²|+c

### Question 20. ∫(x²+1)/(2x+1)(x²-1) dx

Solution:

(x²+1)/(2x+1)(x²-1) =A/(2x+1)+Bx+C/(x²-1)

x²+1=A(x²-1)+(Bx+C)(2x+1)

=(A+2B)x²+(B+2C)*+(-A+c)

Equating similar terms, we get

A+2B=1 , B+2C=0 And -A+C=1

On Solving We get,

A=-5/3 B=4/3 C=-2/3

Thus,

I=-5/3∫dx/(2x+1) +∫(4x/3-2/3)/(x²-1) dx

=-5/3∫dx/(2x+1)+2/3∫2x/(x²-1)dx-2/3∫dx/(x²-1)

=-5/3∫dx/(2x+1)+2/3∫(2x-1)/((x+1)(x-1)) dx

=-5/3∫dx/(2x+1)+2/3∫{(3/2)/(x+1)+(1/2)/(x-1)}dx

I=-(5/6) * log|2x+1|+log|x+1|+1/3log|x-1|+c

### 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| x|-(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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