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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.30 | Set 1
  • Last Updated : 30 Apr, 2021

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

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