# Class 12 RD Sharma Solution – Chapter 19 Indefinite Integrals – Exercise 19.21

• Last Updated : 16 May, 2021

### 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

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