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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.30 | Set 2

  • Last Updated : 30 Apr, 2021

Question 21. ∫dx/(x[6(log⁡x)²+7log⁡x+2])

Solution:

Let⁡ =∫dx/(x[6(log⁡x)²+7log⁡x+2])

=∫1/(x(2log⁡x+1)(3log⁡x+2)) dx

Now,

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



1=A(3logx+2)+B(2logx+1)

Put x=10-1/2

1=(1/2) A

A = 2n Put x = 10-2/3

1=(-1/3)B

B=-3

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

=log|2logx+1|+log|3logx+2|+c

I=log|(2logx+1)/(3logx+2)|+c

Question 22. ∫ 1/x(xn+1) dx

Solution:

1/x(xn+1)

Multiplying numerator and denominator by xn-1, we obtain

1/x(xn+1) =xn-1/(xn-1 x(xn+1))=xn-1/(xn (xn+1))

Let xn=t

x n-1dx=dt

∫ 1/x(xn+1) dx=∫ xn-1/(xn (xn+1)) dx

=1/n ∫ 1/(t(t+1)) dt

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



1=A(1+t)+Bt ——————————-(i)

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

A=1 and B=-1

∫ 1/(t(t+1)) =1/t -1/(1+t)

∫ 1/x(xn+1) dx =1/n ∫ {1/t-1/(t+1)}dx

=1/n[log⁡|t|-log⁡|t+1|]+c

=-1/n [log⁡|xn |-log⁡|xn+1|]+c [xn=t]

=1/n log⁡|xn/(xn+1)|+Cc

Question 23. ∫ x/(x²-a²)(x²-b²) dx

Solution:

Let x/(x²-a²)(x²-b²) =(Ax+B)/(x²-a²) +(Cx+D)/(x²-b²)

x =(Ax+B)(x²-b²)+(Cx+D)(x²-a²)

x=(A+C)x3+(B+D)x²+(-Ab²-Ca²)x+(-Bb²-Da²)

A+C=0,

B+D=0,

-Ab²-Ca²=1,

-Bb²-Da²=0

We get B=0 , D=0 , C=1/(b²-a²) And A=-1/(b²-a²)

Thus,

I=-1/(b²-a²)∫x dx/(x²-a²) + 1/(b²-a²)∫x dx/(x²-b²)

I=-1/2(b²-a²) log⁡|x²-a² |+1/2(b²-a²) log⁡|x²-b² |+c



Question 24.∫(x²+1)/(x²+4)(x²+25) dx

Solution:

Consider the integral I=∫(x²+1)/(x²+4)(x²+25) dx

Let y=x²

Thus,

(x²+1)/(x²+4)(x²+25) =(y+1)/((y+4)(y+25))

(y+1)/((y+4)(y+25))=A/(y+4)+B/(y+25)

(y+1)/((y+4)(y+25)) = (A(y+25)+B(y+4))/((y+4)(y+25))

y+1=Ay+25A+By+4B

Comparing the coefficients, we have,

A+B=1 and 25A+4B=1

Solving the above equations, we have

A=(-1)/7 and B=8/7

Thus, ∫(x²+1)/(x²+4)(x²+25) dx

=∫((-1)/7)/(x²+4) dx+∫(8/7)/(x²+25) dx

=(-1)/7∫1/(x²+4) dx+8/7∫1/(x²+25) dx

=(-1)/7×1/2 tan-1x/2+8/7×1/5 tan-1⁡x/5+C

=(-1)/14 tan-1x/2+8/35 tan-1x/5+C

Question 25. ∫(x3+x+1)/(x²-1) dx

Solution:

Let I =∫(x3+x+1)/(x2-1) dx

=∫(x+(2x+1)/(x²-1))dx

Now,

Let (2x+1)/(x²-1)=A/(x+1)+B/(x-1)

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

Put x=1

3=2B

B=3/2

Put x=-1

-1=-2A

A=1/2

I=∫x dx+1/2∫dx/(x+1)+3/2∫dx/(x-1)



I=x²/2+1/2 log⁡|x+1|+3/2 log⁡|x-1|+c

Question 26. ∫(3x-2)/((x+1)² (x+3)) dx

Solution:

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

3x-2=A(x+1)(x+3)+B(x+3)+C(x+1)²

=(A+C)x²+(4A+B+2C)x+(3A+3B+C)

Equating similar terms, we get,

A+C=0

A=-C

4A+B+2C=3

B=-2C=3

3A+3B+C=-2

3B-2C=-2

Solving, we get,

B=-5/2 , C=-11/4 And A=11/4

Thus,

I=11/4 ∫ dx/(x+1)-5/2 ∫ dx/((x+1)²)-11/4 ∫ dx/(x+3)

I=11/4 log⁡|x+1|+5/(2(x+1))-11/4 log⁡|x+3|+c

Question 27. ∫ (2x+1)/((x+2)(x-3)²) dx

Solution:

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

2x+1=A(x-3)²+B(x+2)(x-3)+C(x+2)

=(A+B)x²+(-6A-B+C)x+(9A-6B+2C))

Equating similar terms, we get,

A+B=0

A=-B

-6A-B+C=2

5B+C=2

9A-6B+2C=1

-15B+2C=1

Solving, we get,

B=3/25 , C=7/5 And A=-3/25

Thus,

I=-3/25 ∫ dx/(x+2)+3/25 ∫ dx/(x-3)+7/5 ∫ dx/((x-3)²)

I=-3/25 log⁡|x+2|+3/25 log⁡|x-3|-7/(5(x-3))+c

Question 28. ∫ (x²+1)/((x-2)² (x+3)) dx

Solution:

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

x²+1=A(x-2)(x+3)+B(x+3)+C(x-2)²

=(A+C)x²+(A+B-4C)x+(-6A+3B+4C)

Equating similar terms, we get,

A+C=1,

A+B-4C=0,

-6A+3B+4C=1

Solving, we get,

A=3/5,

B=1,

C=2/5

Thus,

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

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

Question 29. ∫ x/((x-1)² (x+2)) dx

Solution:

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



x=A(x-1)(x+2)+B(x+2)+C(x-1)²

Substituting x=1, we obtain

B=1/3

Equating the coefficients of x² and constant term, we obtain

A+C=0

-2A+2B+C=0

On solving, we obtain

A=2/9 and C=(-2)/9

x/((x-1)² (x+2)) = 2/(9(x-1))+1/(3(x-1)²)-2/(9(x+2))

∫x/((x-1)² (x+2)) dx =2/9∫1/((x-1)) dx+1/3∫1/((x-1)²) dx-2/9∫1/((x+2)) dx

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

=2/9 log⁡|(x-1)/(x+2)|-1/(3(x-1))+c

Question 30.  ∫ x²/((x-1)(x+1)²) dx

Solution:

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

x²=A(x+1)²+B(x-1)(x+1)+C(x-1)

=(A+B)x²+(2A+C)x+(A-B-C)

Equating similar terms,

A+B=1,

2A+C=0,

A-B-C=0

Solving, we get,

A=1/4,

B=3/4,

C=-1/2

Thus,

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

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

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

Question 31.  ∫ (x²+x-1)/((x+1)² (x+2)) dx

Solution:

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

=(A+C)x²+(3A+B+2C)x+(2A+2B+C))

Equating similar terms

A+C=1 , 3A+B+2C=1 And 2A+2B+C=-1

Solving, we get, A=0 , B=-1 And C=1

Thus,

I =0*∫ dx/(x+1)+(-1)∫ dx/((x+1)²)+1*∫ dx/((x+2))

=+1/(x+1)+log⁡|x+2|+c

I =1/(x+1)+log⁡|x+2|+c

Question 32.  ∫(2x²+7x-3)/(x² (2x+1)) dx

Solution:

Let (2x²+7x-3)/(x² (2x+1))=A/x+B/x² +C/(2x+1)

2x²+7x-3=Ax(2x+1)+B(2x+1)+Cx²

Equating similar terms, we get,

2A+C=2 , A+2B=7 And B=-3

Solving, we get, A=13 And C=-24

Thus,

I= ∫ 13dx/x-∫3dx/x² -24∫dx/(2x+1)

I=13log⁡|x|+3/x-12log⁡|2x+1|+c

Question 33. ∫(5x²+20x+6)/(x3+2x²+x) dx

Solution:

Let I=∫(5x²+20x+6)/(x3+2x²+x) dx



=∫(5x²+20x+6)/(x(x+1)²) dx

Now,

Let

(5x²+20x+6)/(x(x+1)²)=A/x+B/(x+1)+C/((x+1)²)

5x²+20x+6=A(x+1)²+Bx(x+1)+Cx-1

Equating similar terms, we get,

A+B=5 , 2A+B+C=20 And A=6

Solving, we get, B=-1 And C=9

Thus,

I=∫6dx/x-1∫dx/(x+1)+9∫dx/((x+1)²)

I=6log⁡|x|-log⁡|x+1|-9/(x+1)+c

Question 34. ∫ 18/((x+2)(x²+4)) dx

Solution:

Let 18/((x+2)(x²+4))=A/(x+2)+(Bx+C)/(x²+4)

18=A(x²+4)+(8x+c)(x+2)

18=(A+8)x²+(2B+C)x+(4A+2C)

Equating similar terms, we get,

A+B=0 , 2B+C=0 And 4A+2C=18

Solving, we get,

A=9/4, B=-9/4 And C=9/2

Thus,

I=9/4 ∫ dx/(x+2)+(-9/4) ∫ x/(x²+4) dx+9/2 ∫ dx/(x²+4) [∫ dx/(x²+a²)=1/a tan-1x/a]

I=9/4 log⁡|x+2|-9/8 log⁡|x²+4|+9/4 tan-1(x/2)+c [∵∫▒ dx/(x^2+a^2)=1/a tan^(-1)⁡x/9])

Question 35. ∫ 5/((x²+1)(x+2)) dx

Solution:

Let 5/((x²+1)(x+2))=(Ax+B)/(x²+1)+C/(x+2)

5=(Ax+B)(x+2)+C(x²+1)

Equating similar terms, we get,

A+C=0 , 2A+B=0 And 2B+C=5

Solving, we get,

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

Thus,

I =∫ (-x+2)/(x²+1) dx+∫ dx/(x+2)

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

=-1/2 log⁡|x²+1|+2tan-1x+log⁡|x+2|+c

Question 36. ∫x/((x+1)(x²+1))dx

Solution:

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

x=A(x²+1)+(Bx+C)(x+1)

Equating similar terms, we get,

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

Solving, we get, A=-1/2 , B=1/2 And C=1/2

Thus,

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

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

Question 37. ∫dx/(1+x+x²+x3)

Solution:

Let I=∫dx/(1+x+x²+x3)

I=∫dx/((x²+1)(x+1))

Now,

Let 1/((x²+1)(x+1))=(Ax+B)/(x²+1)+C/(x+1)

1=(Ax+B)(x+1)+C(x²+1)

Equating similar terms, we get,

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

Solving, we get,

A=-1/2, B=1/2 And C=1/2

Thus,

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

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

Question 38. ∫ 1/((x+1)²(x²+1)) dx

Solution:

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

1 =A(x+1)(x²+1)+B(x²+1)+(Cx+D)(x+1)²

=(A+C)x3+(A+B+2C+D)x²+(A+C+2D)x+(A+B+D)

Equating similar terms, we get,



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

Solving, we get,

A=1/2, B=1/2, C=-1/2 And D=0

Thus,

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

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

Question 39. ∫2x/(x3-1) dx

Solution:

Let I=∫2x/(x3-1) dx

=∫2x/((x-1)(x²+x+1)) dx

Now,

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

2x=A(x²+x+1)+(Bx+C)(x-1)

=(A+B)x²+(A-B+C)x+(A-C)

Equating similar terms,

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

Solving, we get,

A=2/3, B=-2/3, C=2/3

Thus,

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

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

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

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

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

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

Hence,

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

Question 40. ∫ 1/(x²+1)(x²+4) dx

Solution:

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

1= (Ax+B)(x²+4)+(Cx+D)(x²+1)

=(A+C)x3+(B+D)x²+(4A+C)x+4B+D

Equating similar terms, we get

A+C=0, B+D=0, 4A+C=0 And 4B+D=1

Solving, we get, A=0, B=1/3, C=0, D=-1/3

Thus,

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

=1/3 tan-1⁡x-1/6 tan-1⁡(x/2)+c [∫ dx/(x²+a²)=1/a tan⁡-1x/a]

I=1/3 tan-1⁡x-1/6 tan-1⁡(x/2)+c

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