RD Sharma Class 12 Ex 19.21 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.21 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 19.21 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

RD Sharma Class 12 Ex 19.21 Solutions Chapter 19 Indefinite Integrals

Question 1. ∫x/√(x²+6x+10) dx

Solution:

Let l=∫x/√(x²+6x+10) dx

Let x=2 d/dx (x²+6x+10)+μ

=λ(2x+6)+μ

x=(2λ)x+6λ+μ

Comparing the coefficients of x,

2λ=1

λ=1/2

6λ+μ=0

6(1/2)+μ=0

μ=-3

so, l₁=∫(1/2(2x+6)-3)/√(x²+6x+10) dx

=1/2 ∫ ((2x+6))/√(x²+6x+10) dx-3∣1/√(x²+2x(3)+(3)²-(3)²+10) dx

I₁=1/2 ∫ (2x+6)/√(x²+6x+10) dx-3] 1/√((x+3)²+(1)²) dx

l₁=1/2(2√(x²+6x+10))-3log⁡|x+3+√((x+3)²+1)|+c

[∫1/√x dx=2√x+c, ∫1/√(x²+a²) dx=log⁡|x+√(x²+a²)|+c]

I=√(x²+6x+10)-3log⁡|x+3+√(x²+6x+10)|+c

Question 2. ∫ (2x+1)/√(x²+2x-1) dx

Solution:

Let I= ∫ (2x+1)/√(x²+2x-1) dx

Let 2x +1=λ d/dx (x²+2x-1)+μ

=λ(2x+2)+μ

2x+1=(2λ)x+2λ+μ

Comparing the coefficients of x ,

2λ=2

λ=1

2λ+μ=1

2(1)+μ=1

μ=-1

so, I=∫((2x+2)-1)/√(x²+2x-1) dx

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

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

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

I=2√(x²+2x-1)-log⁡|x+1+√(x²+2x-1)|+c

Question 3. ∫ (x+1)/√(4+5x-x²) dx

Solution:

Let I=∫ (x+1)/√(4+5x-x²) dx

Let x+1=λ d/dx (4+5x-x²)+μ

=λ(5-2x)+μ

x=(-2λ)x+5λ+μ

Comparing the coefficients of x,

so,

-2λ=1

λ=-1/2

5λ+μ=1

5(-1/2)+μ=1

μ=7/2

I=∫ (-1/2(5-2x)+7/2)/√(4+5x-x²) dx

=-1/2 ∫ ((5-2x))/√(4+5x-x²) dx+7/2 ∫ 1/√(-[x²-5x-4]) dx

I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√(-[x²-2x(5/2)+(5/2)²-(5/2)²-4]) dx

I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√(-[(x-5/2)²-(√41/2)²]) dx

I=-1/2 ∫ (5-2x)/√(4+5x-x²) dx+7/2 ∫ 1/√((√41/2)²-(x-5/2)²]) dx

I=-1/2 (2√(4+5x-x²))+7/2 sin^-1((x-5/2)/(√41/2))+c

I=-√(4+5x-x²)+7/2 sin^-1⁡((2x-5)/√41)+c

Question 4. ∫(6x-5)/√(3x²-5x+1) dx

Solution:

Let I=∫(6x-5)/√(3x²-5x+1) dx

Let 3x²-5x+1=t

(6x-5)dx=dt

I=∫ dt/√t

=2√t+c

I=2√(3x²-5x+1)+c

Question 5. ∫(3x+1)/√(5-2x-x²) dx

Solution:

Let I=∫(3x+1)/√(5-2x-x²) dx

Let 3x+1=λ d/dx (5-2x-x²)+μ

=λ(-2-2x)+μ

3x+1=(-2λ)×-2λ+μ

Comparing the coefficients of x,

-2λ=3

λ=-3/2

-2λ+μ=1

-2(-3/2)+μ=1

μ=-2

so, I=∫ (-3/2(-2-2x)-2)/√(5-2x-x²) dx

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

I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x+(1)²-(1)²-5]) dx

I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[(x+1)²-(√6)²]) dx

[∫ 1/√x dx=2√x+c,∫ 1/√(a²-x²) dx=sin-1⁡(x/a)+c]

I=-3/2×2√(5-2x-x²)-2sin^-1⁡((x+1)/√6)+c

Question 6. ∫ x/√(8+x-x²) dx

Solution:

Let I =∫ x/√(8+x-x²) dx

Let x =λ d/dx (8+x-x²)+μ

=λ(1-2x)+μ

x=(-2λ)x+λ+μ

Comparing the coefficients of x,

so,

-2λ=1

λ=-1/2

λ+μ=0

(-1/2)+μ=0

μ=1/2

I=∫ (-1/2(1-2x)+1/2)/√(8+x-x²) dx

=-1/2 ∫ ((1-2x))/√(8+x-x²) dx+1/2 ∫ 1/√(-[x²-x-8]) dx

I=-1/2 ∫ (1-2x)/√(8+x-x²) dx+1/2 ∫ 1/(√(-[x²-2x(1/2)+(1/2)²-(1/2)²-8]) dx

I=-1/2 ∫ (1-2x)/√(8+x-x²) dx+1/2 ∫ 1/(√(-[(x-1/2)²-(33/4)²]) dx

I=-1/2 ∫ ((1-2x))/√(8+x-x²) dx+1/2 ∫ 1/√((√33/2)²-(x-1/2)²]) dx

I=-1/2×2√(8+x-x²)+1/2 sin^-1⁡((x-1/2)/(√33/2))+c

I=-√(8+x-x²)+1/2 sin^-1⁡((2x-1)/√33)+c

Question 7. ∫(x+2)/√(x²+2x-1) dx

Solution:

Let l=∫(x+2)/√(x²+2x-1) dx

Let x+2=λ d/dx (x²+2x-1)+μ

x+2=λ(2x+2)+μ

x+2=(2λ)x+2λ+μ

Comparing the coefficients of x,

2λ =1 ∫

λ =1/2

2λ +μ=2

2*(1/2)+μ=2

μ=1

So, I = ∫(1/2(2x+2)+1)dx/√(x²+2x-1)

=1/2∫(2x+2)dx/√(x²+2x-1) + ∫1dx//√(x²+2x+1²-1²-1)

=1/2∫(2x+2)dx/√(x²+2x-1) + ∫1dx/(√(x+1)²-√2²)

I=1/2(2*(x²+2x-1)) +log|x+1+(√(x+)²-√2²)|+c

I=(x²+2x-1) +log|x+1+√(²+2x+1)|+c

Question 8. ∫(x+2)/√(x²-1) dx

Solution:

Let x+2=A d/dx (x²-1)+B——————————-(i)

x+2=A(2x)+B

Equating the coefficients of x and constant term on both sides, we obtain

2A=1

A=1/2

B=2

From (1), we obtain

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

Then, ∫ (x+2)/√(x²-1) dx

=∫ (1/2(2x)+2)/√(x²-1) dx

=1/2 ∫ 2x/√(x²-1) dx+∫ 2/√(x²-1) dx———————–(ii)

In⁡1/2 ∫ 2x/√(x²-1) dx,

let x²-1=t

2xdx=dt

1/2 ∫ 2x/√(x²-1) dx

=1/2 ∫ dt/√t

=1/2[2√t]

=√t

=√(x²-1)

Then, ∫2/√(x²-1) dx=2∫1/√(x²-1) dx=2log∣x+√(x²-1)

From equation (2), we obtain

∫(x+2)/√(x²-1) dx=√(x²-1)+2log⁡|x+√(x²-1)|+c

Question 9. ∫ (x-1)/√(x²+1) dx

Solution:

∫ (x-1)/√(x²+1) dx

=∫ (x-1)/√(x²+1) dx

=∫ x/√(x²+1)-1/√(x²+1) dx

=1/2 ∫ 2x/√(x²+1) dx-∫ 1/√(x²+1) dx [let t=x²+1, dt=2xdx]

=1/2 ∫ dt/√t-∫ 1/√(x²+1) dx

=1/2(2√t)-∫ 1/√(x²+1) dx

=√t-ln⁡|x+√(x²+1)|+c

=√(x²+1)-ln⁡|x+√(x²+1)|+c

Question 10. ∫ x/√(x²+x+1) dx

Solution:

Let I =∫ x/√(x²+x+1) dx

Let x =λ d/dx (x²+x+1)+μ

=λ(2x+1)+μ

x=(2λ)×+λ+μ

Comparing the coefficients of x,

So,

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

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

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

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

Question 11. ∫ (x+1)/√(x²+1) dx

Solution:

Let I=∫ (x+1)/√(x²+1) dx

Let x+1=λ d/dx (x²+1)+μ

x+1=λ(2x)+μ

Comparing the coefficients of x,

2λ=1

λ=1/2

μ=1

I=∫ (1/2(2x)+1)/√(x²+1) dx

=1/2 ∫ (2x)/√(x²+1) dx+∫ 1/√(x²+1) dx

I=1/2×2√(x²+1)+log⁡|x+√(x²+1)|+c

I=√(x²+1)+log⁡|x+√(x²+1)|+c

Question 12. ∫ (2x+5)/√(x²+2x+5) dx

Solution:

Let I= ∫ (2x+5)/√(x²+2x+5) dx

Let 2x +5=λ d/dx (x²+2x+5)+μ

=λ(2x+2)+μ

2x+5=(2λ)x+2λ+μ

Comparing the coefficients of x,

2λ=2

λ=1

2λ+μ=5

2(1)+μ=5

μ=3

so, l =∫ ((2x+2)+3)/√(x²+2x+5) dx

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

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

I =2√(x²+2x+5)+3log⁡|x+1+√((x+1)²+(2)²)|+c

I=2√(x²+2x+5)+3log⁡|x+1+√(x²+2x+5)|+c

Question 13. ∫ (3x+1)/√(5-2x-x²) dx

Solution:

Let I= ∫ (3x+1)/√(5-2x-x²) dx

Let 3x +1=λ d/dx (5-2x-x²)+μ

=λ(-2-2x)+μ

3x+1=(-2λ)x-2λ+μ

Comparing the coefficients of x ,

-2λ=3

λ=-3/2

-2λ+μ=1

-2(-3/2)+μ=1

μ=-2

Now, I=∫ (-3/2(-2-2x)-2)/√(5-2x-x²) dx

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

I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[x²+2x+(1)²-(1)²+5]) dx

I=-3/2 ∫ (-2-2x)/√(5-2x-x²) dx-2∫ 1/√(-[(x+1)²-(√6)²]) dx

I=-3/2×2√(5-2x-x²)-2sin^-1⁡((x+1)/√6)+c

I=-3√(5-2x-x²)-2sin^-1((x+1)/√6)+c

Question 14. ∫ √((1-x)/(1+x)) dx

Solution:

Let I=∫ √((1-x)/(1+x)) dx

=∫ √((1-x)/(1+x)×(1-x)/(1-x)) dx

=∫ (1-x)/√(1-x²) dx

Let 1-x=λ d/dx (1-x²)+μ

=λ(-2x)+μ

1-x=(-2λ)x+μ

Comparing the coefficients of x,

-2λ=-1

λ=1/2

μ=1

I=∫ (1/2(-2x)+1)/√(1-x²) dx

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

I=1/2×2√(1-x²)+sin^-1⁡x+c

I=√(1-x²)+sin^-1⁡x+c

Question 15. ∫ (2x+1)/√(x²+4x+3) dx

Solution:

Let I= ∫ (2x+1)/√(x²+4x+3) dx

Let 2x +1=λ d/dx (x²+4x+3)+μ

=λ(2x+4)+μ

2x+1=(2λ)x+4λ+μ

Comparing the coefficients of x,

2λ=2

λ=1

4λ+μ=1

4*1+μ=1

μ=-3

I=∫((2x+4)-3)/√(x²+4x+3) dx

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

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

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

I=2√(x²+4x+3)-3log⁡|x+2+√(x²+4x+3)|+c

Question 16. ∫ (2x+3)/√(x²+4x+5) dx

Solution:

Let I=∫ (2x+3)/√(x²+4x+5) dx

Let 2x+3=λ d/dx (x²+4x+5)+μ

=λ(2x+4)+μ

2x+3=(2λ)x+4λ+μ

Comparing the coefficients of x,

2λ=2

λ=1

4λ+μ=3

4(1)+μ=3

μ=-1

Now, I=∫ ((2x+4)-1)/√(x²+4x+5) dx

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

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

I=2√(x²+4x+5)-log⁡|x+2+√((x+2)²+1)|+c

I=2√(x²+4x+5)-log⁡|x+2+√(x²+4x+5)|+c

Question 17. ∫(5x+3)/√(x²+4x+10) dx

Solution:

∫(5x+3)/√(x²+4x+10) dx

let 5x+3=λ(2x+4)+μ

λ=5/2,

μ=-7

∫(λ(2x+4)+μ)/√(x²+4x+10) dx

=∫(5/2(2x+4)-7)/√(x²+4x+10) dx

=∫(5/2(2x+4))/√(x²+4x+10) dx-∫7/√(x²+4x+10) dx

=∫(5/2 dt)/√t-∫7/√((x+2)²+6) dx

=5√(x²+4x+10)-7log⁡|(x+2)+√(x²+4x+10)|+c

Question 18. ∫(x+2)/√(x²+2x+3) dx

Solution:

Let I=∫(x+2)/√(x²+2x+3) dx

x+2=A d/dx [x²+2x+3]+B

x+2=2Ax+2A+B

Comparing the co-efficients, we have, 2A=1 and 2A+B=2

A=1/2

Substituting the value of A in 2A+B=2, we have, 2×1/2+B=2

1+B=2

B=2-1

B=1

Thus we have, x+2=1/2[2x+2]+1

Hence, I=∫(x+2)/√(x²+2x+3) dx

=∫[1/2[2x+2]+1]/√(x²+2x+3) dx

=∫[1/2[2x+2]+1]/√(x²+2x+3) dx

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

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

Substituting t=x²+2x+3 and dt=2x+2 in the first integrand,

we have, I=1/2∫dt/√t+∫dx/√(x²+2x+3)

=1/2×2√t+∫dx/√(x²+2x+1+2)+c

=√t+∫dx/√((x+1)²+(√2)²)+c

I=√(x²+2x+3)+log⁡[|x+1|+√((x+1)²+(√2)²)]+c

I=√(x²+2x+3)+log⁡[|x+1|+√(x²+2x+3)]+c

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.