RD Sharma Class 12 Ex 19.30 Solutions Chapter 19 Indefinite Integrals

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.30
CategoryRD Sharma Solutions

Table of Contents

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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