# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.19

• Last Updated : 04 May, 2021

### Question 1. ∫ x/(x2 + 3x + 2) dx

Solution:

Given that I = ∫ x/(x2 + 3x + 2) dx

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12.

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Let x = m d/dx (x2 + 3x + 2) + n

= m(2x + 3) + n

x = (2m)x + (3λ + n)

On comparing the coefficients of x,

2m = 1

m = 1/2

3m + n = 0

3(1/2) + n = 0

n = -3/2

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

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

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

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

= 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 3/2 – 1/2)/(x + 3/2 + 1/2)| + c

As we know that ∫1/(a2 – x2)dx = 1/2a log|(x – a)/(x + a)| + c]

Hence, I = 1/2log|x2 + 3x + 2| – (3/2) × (1/2 × (1/2))log|(x + 1)/(x + 2)| + c

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

Solution:

Given that I = ∫(x + 1)/(x2 + x + 3) dx

Let us considered x + 1 = m d/dx(x2 + x + 3) + n

x + 1 = m(2x + 1) + n

x + 1 = (2m)x + (m + n)

On comparing the co-efficient of x,

2m = 1

m = 1/2

m + n = 1

(1/2) + n = 1

n = 1/2

Now,

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

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

= 1/2∫(2x + 1)/(x2 + x + 1) dx + 1/2∫1/((x + 1/2)2 + (11/4)) dx

= 1/2 ∫(2x + 1)/(x2 + x + 1) dx + 1/2 ∫1/((x + 1/2)2 + (√11/2)2) dx

= 1/2 log|x2 + x + 3| + (1/2) * (1/(√11/2)) tan-1((x + 1/2)/(√11/2)) + c

As we know that ∫1/(x2 + a2) dx = 1/a tan-1(x/a) + c

Hence, I = 1/2 log|x2 + x + 3| + 1/√11 tan-1((2x + 1)/(√11)) + c

### Question 3. ∫ (x – 3)/(x2 + 2x – 4) dx

Solution:

Given that I = ∫(x – 3)/(x2 + 2x – 4) dx

Let us considered x – 3 = m d/dx (x2 + 2x – 4) + n

= m(2x + 2) + n

x – 3 = (2m)x + (2m + n)

On comparing the coefficients of x,

2m = 1

m = 1/2

2m + n = -3

2(1/2) + n = -3

n = -4

So,

I = ∫(1/2(2x + 2) – 4)/(x2 + 2x – 4) dx

= 1/2 ∫(2x + 2)/(x2 + 2x – 4) dx – 4∫ 1/(x2 + 2x + (1)2 – (1)2 – 4) dx

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

As we know that ∫1/(x2 – a2) dx = 1/2a log|(x – a)/(x + a)| + c

= 1/2 log⁡|x2 + 2x – 4| – 4 × 1/(2√5) log⁡|(x + 1 – √5)/(x + 1 + √5)| + c

Hence, I = 1/2 log⁡|x2 + 2x – 4| – 2/√5 log⁡|(x + 1 – √5)/(x + 1 + √5)| + c

### Question 4. ∫(2x – 3)/(x2 + 6x + 13) dx

Solution:

Given that I = ∫ (2x – 3)/(x2 + 6x + 13) dx

Let us considered 2x – 3 = m d/dx (x2 + 6x + 13) + n

= m(2x + 6) + n

2x – 3 = (2m)x + (6m + n)

On comparing the co-efficient of x, we get

2m = 2

m = 1

6m + n = -3

6 * 1 + n = -3

n = -9

Now,

= ∫(1 * (2x + 6) – 9)/(x2 + 6x + 13) dx

= ∫(2x + 6)/(x2 + 6x + 13) dx + ∫(-9)/(x2 + 2 * (3) * x + (3)2 – (3)2 + 13) dx

= ∫(2x + 6)/(x2 + 6x + 13) dx -9 ∫1/((x + 3)2 + (2)) dx

= log|(x2 + 6x + 13)| – 9 * (1/2) tan-1((x + 3)/2) + c

Hence, I = log|(x2 + 6x + 13)| – 9 × (1/2) tan-1((x + 3)/2) + c

### Question 5. ∫x2/(x2 + 7x + 10) dx

Solution:

Given that I = ∫x2/(x2 + 7x + 10) dx

= ∫{1 – (7x + 10)/(x2 + 7x + 10)}dx

I = x – ∫(7x + 10)/(x2 + 7x + 10) dx + c1   ……..(i)

Let I1 = ∫(7x + 10)/(x2 + 7x + 10) dx

Let 7x + 10 = md/dx (x2 + 7x + 10) + n

= m(2x + 7) + n

7x + 10 = (2m)x + 7m + n

On comparing the coefficients of like powers of x,

7 = 2m

m = 7/2

7m + n = 10

7(7/2) + n = 10

n = -29/2

So, l = ∫(1/6(6x – 4) – 1/3)/(3x2 – 4x + 3) dx

= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 4/3x + 1) dx

= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x2 – 2x(2/3) + (2/3)2 – (2/3)2 + (2)2) dx

= 1/6 ∫(6x – 4)/(3x2 – 4x + 3) dx – 1/9 ∫1/(x – 2/3)2 + (√5/2)) dx

= 1/6log|(3x2 – 4x + 3) | – ((1/9) × 1/(√5/3)) tan((x – 2/3)/(√5/3)) + c

Hence, I = 1/6log|(3x2 – 4x + 3)| – (√5/15)tan-1((3x – 2)/√5) + c

### Question 6. ∫2x/(2 + x – x2) dx

Solution:

Given that I = ∫2x/(2 + x – x2) dx

Now,

2x = m(d/dx(2 + x + x2)) + n

2x = m(-2x + 1) + n

Now equating the co-efficient of we will get m, n

m = -1,

n = 1

∫2x/(2 + x – x2) dx

= ∫(m(-2x + 1) + n)/(2 + x – x2) dx

= ∫(-1(-2x + 1) + 1)/(2 + x – x2) dx

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

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

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

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

= -log⁡|2 + x – x2| + ∫1/((x – 1/2)2 – (3/2)2) dx

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

Hence, I = -log⁡|2 + x – x2| – 1/3 log⁡|(x – 2)/(x + 1)| + c

### Question 7. ∫(1 – 3x)/(3x2 + 4x + 2) dx

Solution:

Given that I = ∫(1 – 3x)/(3x2 + 4x + 2) dx

Let us considered 1 – 3x = m d/dx (3x2 + 4x + 2) + n

= m(6x + 4) + n – 11 – 3x = (6m)× + (4λ + n)

On comparing the coefficients of l x,

6m = -3

m = -1/2

4m + n = 1

4(-1/2) + n = 1

n = 3

I = ∫(-1/2(6x + 4) + 3)/(3x2 + 4x + 2) dx

= -1/2 ∫ (6x + 4)/(3x2 + 4x + 2) dx + 3∫1/(3x2 + 4x + 2) dx

= -1/2 ∫   (6x + 4)/(3x2 + 4x + 2) dx + 3/3 ∫1/(x2 + 4/3x + 2/3) dx

= -1/2 ∫   (6x + 4)/(3x2 + 4x + 2) dx + ∫1/(x2 + 2x(2/3) + (2/3)2 – (2/3)2 + (2/3) dx

= -1/2 ∫   (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + 2/9) dx

= -1/2 ∫   (6x + 4)/(3x2 + 4x + 2) dx + ∫1/((x + 2/3)2 + (√2/3)2) dx

= -1/2 log|(3x2 + 4x + 2) | + 3/√2tan-1((x + 2/3)/(√2/3)) + c

Hence, I = -1/2 log|(3x2 + 4x + 2)| + 3/√2tan-1((3x + 2)/√2) + c

### Question 8. ∫(2x + 5)/(x2 – x – 2) dx

Solution:

Given that I = ∫(2x + 5)/(x2 – x – 2) dx

Let 2x + 5 = md/dx (x2 – x – 2) + n

= m(2x – 1) + n

2x + 5 = (2m)x – m + n

On comparing the coefficients of x,

2m = 2

m = 1

-m + n = 5

-1 + n = 5

n = 6

So,

I = ∫((2x – 1) + 6)/(x2 – x – 2) dx

= ∫ ((2x – 1))/(x2 – x – 2) dx + 6∫1/(x2 – 2x(1/2) + (1/2)2 – (1/2)2 – 2) dx

= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – 9/4) dx

= ∫ (2x – 1)/(x2 – x – 2) dx + 6∫1/((x – 1/2)2 – (3/2)2) dx

= log⁡|x2 – x – 2| + 6/2(3/2) log⁡|(x – 1/2 – 3/2)/(x – 1/2 + 3/2)| + c

As we know that ∫1/(x2 – a2) dx = 1/2a log⁡|(x – a)/(x + a)| + c

Hence, I = log⁡|x2 – x – 2| + 2log⁡|(x – 2)/(x + 1)| + c

### Question 9. ∫ (ax3 + bx)/(x4 + c2) dx

Solution:

Given that I = ∫(ax3 + bx)/(x4 + c2) dx

Let us considered ax3 + bx = md/dx (x4 + c2) + n

ax3 + bx = n(4x3) + n

On comparing the coefficients of x,

4m = a

m = a/4

and

n = 0

I = ∫(a/4 (4x3) + bx)/(x4 + c2) dx

= a/4 ∫(4x3)/(x4 + c2) dx + b∫x/((x2)2 + c2) dx

= a/4 ∫(4x3)/(x4 + c2) dx + b/2 ∫2x/((x2)2 + c2) dx

= a/4 log⁡|x4 + c2| + b/2 I1   ……..(i)

Now,

I1 = ∫2x/((x2)2 + c2) dx

Put x2 = t

2xdx = dt

I1 = ∫1/((t)2 + c2) dx

= 1/c tan-1⁡(t/c) + c1

l1 = 1/C tan-1⁡(x2/c) + c1        …………(ii)

Now using equation (i) and (ii) we get,

Hence, I = a/4 log|x4 + c2| + b/2c tan(x2/c) + b

### Question 10. ∫(x + 2)/(2x2 + 6x + 5) dx

Solution:

Given that I = ∫(x + 2)/(2x2 + 6x + 5) dx

Let us considered x + 2 = m d/dx (2x2 + 6x + 5) + n

= m(4x + 6) + n

x + 2 = (4m)x + (6m + n)

On comparing the coefficients of x,

So,

4m = 1

m = 1/4

6m + n = 2

6(1/4) + n = 2

n = 1/2

I = ∫(1/4(4x + 6)+1/2)/(2x2 + 6x + 5) dx)

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/2 ∫1/(2x2 + 6x + 5) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 3x + 5/2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/(x2 + 2x(3/2) + (3/2)2 – (3/2)2 + 5/2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + 1/4) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 ∫1/((x + 3/2)2 + (1/2)2) dx

=1/4 ∫(4x + 6)/(2x2 + 6x + 5) dx + 1/4 × (1/(1/2))tan-1⁡((x + 3/2)/(1/2)) + c

As we know that ∫1/(x2 + a2) dx = 1/a tan-1⁡(x/a) + c

Hence, I = 1/4 log|2x2 + 6x + 5| + 1/2 tan-1(2x + 3) + c

### Question 11. ∫((3sin⁡x – 2)cos⁡x)/(5 – cos2⁡x – 4sin⁡x) dx

Solution:

Given that I = ∫((3sin⁡x – 2)cos⁡x)/(5 – cos2x – 4sin⁡x) dx

= ∫((3sin⁡x – 2)cos⁡x)/(5 – (1 – sin2⁡x) – 4sin⁡x) dx

= ∫((3sin⁡x – 2)sin⁡x)/(5 – 1 + sin2x – 4sin⁡x) dx)

Now substitute sin⁡x = t in the above equation

cos⁡xdx = dt

So,

I = ∫(3t – 2)/(4 + t2 – 4t) dt

= ∫((3t – 2))/(t2 – 4t + 4) dt

= ∫(3t – 2)/(t – 2)2 dt

Now Integrate partial fractions.

(3t – 2)/((t – 2)2) = A/((t – 2)) + B/((t – 2)2)

= (A(t – 2) + B)/((t – 2)2)

= (At – 2A + B)/((t – 2)2)

3t – 2 = At – 2A + B

On comparing the coefficients, we have, A = 3

and -2A + B = -2

Now, on substituting the value of A = 3 in the above equation,

-2 × 3 + B = -2

-6 + B = -2

B = 6 – 2

B = 4

So, I = ∫(3t – 2)/(t – 2)2dt

= ∫(3t – 2)/(t – 2) dt + ∫4/(t – 2)2 dt

= 3log|t – 2| – 4(1/(t – 2)) + c

= 3log|t – 2| – 4(1/(t – 2)) + c

Now put the value of t = sinx, we have ,

I = 3log|sinx – 2| – 4(1/(sinx – 2)) + c

### Question 12. ∫(5x – 2)/(1 + 2x + 3x2) dx

Solution:

Given that I = ∫(5x – 2)/(1 + 2x + 3x2) dx

Let us considered 5x – 2 = A d/dx (1 + 2x + 3x2) + B

5x – 2 = A(2 + 6x) + B

5x – 2 = 6 × A + 2A + B

On comparing the Co-efficient  we have, 6A = 5 and 2A + B = -2

A = 5/6

On substituting the value of A in 2A + B = -2, we n have,

2 * 5/6 + B = -2

10/6 + B = -2

B = -2 – 10/6

B = (-12 – 10)/6

B = (-22)/6

B = (-11)/3

5x – 2 = 5/6(2 + 6x) – 11/3

So, I = ∫(5x – 2)/(1 + 2x + 3x2) dx becomes,

I = ∫[5/6(2 + 6x) – 11/3]/(3x2 + 2x + 1) dx

= 5/6∫(2 + 6x)/(3x2+ 2x + 1) dx – 11/3∫dx/(3x2 + 2x + 1)

= 5/6 log⁡(3x2 + 2x + 1) – 11/(3 × 3)∫dx/(x2 + 2/3x + 1/3) + c

= 5/6 log⁡(3x2 + 2x + 1) – 11/9∫dx/(x2 + 2/3 x + (4/3)2 + 1/3 – (4/3)2) + c

= 5/6 log⁡(3x2 + 2x + 1) – 11/9∫dx/((x + 1/3)2 +(√2/3)2) + c

= 5/6 log⁡(3x2 + 2x + 1) – 11/9 × 1/(√2/3) tan-1(((x + 1/3)/(√2/3))] + C

Hence, I = 5/6 log⁡(3x2 + 2x + 1) – 11/(3√2) tan-1⁡[(3x + 1)/√2] + C

My Personal Notes arrow_drop_up