# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.20

• Last Updated : 30 Apr, 2021

### Question 1. Evaluate: ∫(x2 + x + 1)/(x2 – x) dx

Solution:

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

= ∫ [1 + (2x + 1)/(x2 – x)]dx

= x + ∫(2x + 1)/(x2 – x) dx + c

= x + I1 + c1           …….(i)

Now, I1 = ∫(2x + 1)/(x2 – x) dx

Let 2x + 1 = λ d/dx (x2 – x) + μ = λ(2x – 1) + μ

2x + 1 = (2λ)x – λ + μ

By comparing the coefficients of x, we get

2 = 2λ ⇒ λ = 1

-λ + μ = 1 ⇒ μ = 2

I1 = ∫ ((2x – 1) + 2)/(x2 – x) dx)

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

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

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

= log⁡|x2 – x| +

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

So, I1 = log⁡|x2 – x| + 2log⁡|(x – 1)/x| + c2     ……(ii)

Now put the value of I1 in eq(i), we get

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

### Question 2. ∫ (x2 + x – 1)/(x2 + x – 6) dx

Solution:

Given that I = ∫ (x2 + x – 1)/(x2 + x – 6) dx

= ∫[1 + 5 /(x2 + x – 6)]dx

I = x + ∫5/(x2 + x – 6) dx + c1

Let us assume I1 = 5∫1/(x2 + x – 6) dx

I = x + I1 + c1     …..(i)

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

= 5∫ 1/((x + 1/2)2 – (5/2)2dx

= 5 1/2(5/2) log⁡|(x + 1/2 – 5/2)/(x + 1/2 + 5/2)| + c2

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

So, we get

I1 = log⁡|(x – 2)/(x + 3)| + c2      …….(ii)

Now put the value of I1 in eq(i), we get

I = x + log⁡|(x – 2)/(x + 3)| + c

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

Solution:

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

= ∫ (1 – x2)/(x – 2x2) dx

= ∫ (x2 – 1)/(2x2 – x) dx

= ∫ [1/2 + (x/2 – 1)/(2x2 – x)]dx

I = 1/2x + ∫(x/2 – 1)/(2x2 – x) dx + c1

Let us assume I1 = ∫(x/2 – 1)/(2x2 – x) dx

So, I = 1/2x + I1 + c1     …..(i)

Now, let x/2 -1 = λ d/dx (2x2 – x) + μ = λ(4x – 1) + μ

x/2 – 1 = (4λ)x – λ + μ

By comparing the coefficients of x, we get

1/2 = 4λ ⇒ λ = 1/8

-λ + μ = -1 ⇒ -(1/8) + μ = -1

μ = -7/8

I1 = ∫ (1/8(4x – 1) – 7/8)/(2x2 – x) dx

= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/8 ∫1/2(x2 – x/2)dx

= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/16 ∫1/(x2 – 2x(1/4) + (1/4)2 – (1/4)2) dx

= 1/8 ∫(4x – 1)/(2x2 – x) dx – 7/16 [1/((x – 1/4)2 – (1/4)2) dx

= 1/8 log⁡|2x2 – x| – 7/16 × 1/2(1/4) log⁡|(x – 1/4 – 1/4)/(x – 1/4 + 1/4)| + c2

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

So, we get

I1 = 1/8 log⁡|x| + 1/8 log⁡|2x – 1| – 7/8 log⁡|1 – 2x| + 7/8 log⁡2 + 7/8 log⁡|x| + c2

I1 = log⁡|x| – 3/4 log⁡|1 – 2x| + c3       [Here, c3 = c2 + 7/8 log⁡2]

Now put the value of I1 in eq(i), we get

I = 1/2x + log⁡|x| – 3/4 log⁡|1 – 2x| + c

### Question 4. ∫(x2 + 1)/(x2 – 5x + 6)dx

Solution:

Given that I = ∫(x2 + 1)/(x2 – 5x + 6)dx

Now we convert I into proper rational function by dividing x2 + 1 by x2 – 5x + 6

So,

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

Let

(5x – 5)/((x – 2)(x – 3)) = A/(x – 2) + B/(x – 3)

So, we get A + B = 5 and 3A + 2B = 5

On solving both the equations we get A = -5 and B = 10

So,

Hence, ∫(x2 + 1)/((x + 1)2 (x + 3)) dx =∫dx – 5∫1/(x – 2) dx + 10∫x/(x – 3)

I = x – 5log⁡|x – 2| + 10log⁡|x – 3| + 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

Let us assume I1 = ∫(7x + 10)/(x2 + 7x + 10) dx

So, I = x – I1 + c1     …..(i)

Now let us assume 7x + 10 = λ d/dx (x2 + 7x + 10) + μ = λ(2x + 7) + μ

7x + 10 = (2λ)x + 7λ + μ

By comparing the coefficients of x, we get

7 = 2λ ⇒ λ = 7/2

7λ + μ = 10 ⇒ 7(7/2) + μ = 10μ = -29/2

So, I1 = ∫(7/2(2x + 7) – 29/2)/(x2 + 7x + 10) dx

= 7/2 ∫((2x + 7))/(x2 + 7x + 10) dx – 29/2∫1/(x2 + 2x(7/2) + (7/2)2 – (7/2)2 + 10) dx

= 7/2 ∫(2x + 7)/(x2 + 7x + 10) dx – 29/2 {1/((x + 7/2)2 – (3/2)2) dx

= 7/2 log⁡|x2 + 7x + 10| – 29/2 × 1/2(3/2) log⁡|(x + 7/2 – 3/2)/(x + 7/2 + 3/2)| + c2

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

So, we get

I1 = 7/2 log⁡|x2 + 7x + 10| – 29/6 log⁡|(x + 2)/(x + 5)| + c2

Now put the value of I1 in eq(i), we get

I = x – 7/2 log⁡|x2 + 7x + 10| + 29/6 log⁡|(x + 2)/(x + 5)| + c

### Question 6. ∫(x2 + x + 1)/(x2 – x + 1) dx

Solution:

Given that l = ∫(x2 + x + 1)/(x2 – x + 1) dx

= ∫[1 + 2x/(x2 – x + 1)]dx

= x + ∫2x/(x2 – x + 1) dx + c1

Let us assume I1 = ∫2x/(x2 – x + 1) dx

So, I = x + I1 + c1     …..(i)

Now let 2x = λ d/dx (x2 – x + 1) + μ = λ(2x – 1) + μ

2x = (2λ) × -λ + μ

By comparing the coefficients of x, we get

2 = 2λ ⇒ λ = 1

-λ + μ = 0 ⇒ -1 + μ = 0

μ = 1

So, I1 = ∫((2x – 1) + 1)/(x2 – x + 1) dx

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

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

= log⁡|x2 – x + 1| + 2/√3 tan-1⁡((x – 1/2)/(√3/2)) + c2

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

So, we get

I1 = log⁡|x2 – x + 1| + 2/√3 tan-1⁡((2x – 1)/√3) + c2

Now put the value of I1 in eq(i), we get

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

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

Solution:

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

= ∫(x2 – 2x + 1)/(x2 + 2x + 2) dx

= ∫[1 – (4x + 1)/(x2 + 2x + 2)]dx

= x – ∫(4x + 1)/(x2 + 2x + 2) dx + c1

Let us assume l1 = ∫(4x + 1)/(x2 + 2x + 2) dx

So, I = x – I1 + c1     …..(i)

Now, let 4x + 1 = λ d/dx (x2 + 2x + 2) + μ

= λ(2x + 2) + μ = (2k)x + (2λ + μ)

By comparing the coefficients of x, we get

4 = 2λ ⇒ λ = 2

2λ + μ = 1 ⇒ 2(2) + μ = 1

μ = -3

l1 = ∫ (2(2x + 2) – 3)/(x2 + 2x + 2) dx

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

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

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

So, we get

l1 = 2log⁡|x2 + 2x + 2| – 3tan-1⁡(x + 1) + c2

Now put the value of I1 in eq(i), we get

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

### Question 8. ∫(x3 + x2 + 2x + 1)/(x2 – x + 1) dx

Solution:

Given that, I = ∫(x3 + x2 + 2x + 1)/(x2 – x + 1) dx

= ∫[x + 2 + (3x – 1)/(x2 – x + 1)]dx

= x2/2 + 2x + ∫(3x. -1)/(x2 – x + 1) dx + c1

Let us assume l1 = ∫(3x – 1)/(x2 – x + 1) dx

So, I = x2/2 + 2x + I1 + c1     …..(i)

Now, let 3x – 1 = λ d/dx (x2 – x + 1) + μ = λ(2x – 1) + μ

3x – 1 = (2λ)x – λ + μ

By comparing the coefficients of x, we get

3 = 2λ ⇒ λ = 3/2

-λ + μ = -1 ⇒ -(3/2) + μ = -1

μ = 1/2

So, I1 = ∫(3/2(2x – 1) + 1/2)/(x2 – x + 1) dx

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

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

= 3/2 log⁡|x2 – x + 1| + 1/2 × 2/√3 tan-1⁡((x + 1/2)/(√3/2)) + c2

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

So, we get

I1 = 3/2 log⁡|x2 – x + 1| + 1/√3 tan-1⁡((2x + 1)/√3) + c2

Now put the value of I1 in eq(i), we get

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

### Question 9. ∫(x2 (x4 + 4))/(x2 + 4) dx

Solution:

Given that, I = ∫(x2 (x4 + 4))/((x2 + 4)) dx

= ∫ (x6 + 4x2)/((x2 + 4)) dx

= ∫ [x4 – 4x2 + 20 – 80/(x2 + 4)]dx

= x5/5 – (4x3)/3 + 20x – 80 ∫1/(x2 + 4) dx + c1

Let us assume I1 = ∫1/(x2 + 4) dx

So, I = x5/5 – (4x3)/3 + 20x – 80I1 + c1     …..(i)

Now, I1 = ∫1/(x2 + (2)2) dx

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

So, we get

I1 = 1/2 tan-1⁡(x/2) + c

Now put the value of I1 in eq(i), we get

I = x5/5 – (4x3)/3 + 20x – 80/2 tan-1⁡(x/2) + c

I = x5/5 – (4x3)/3 + 20x – 40tan-1(x/2) + c

### Question 10. ∫ x2/(x2 + 6x + 12) dx

Solution:

Given that, l = ∫ x2/(x2 + 6x + 12) dx

= ∫ [1 – (6x + 12)/(x2 + 6x + 12)]dx

= x – ∫(6x + 12)/(x2 + 6x + 12) dx + c1

Let us assume I1 = ∫(6x + 12)/(x2 + 6x + 12) dx

So, I = x – I1 + c1     …..(i)

Now, let 6x + 12 = λ d/dx (x2 + 6x + 12) + μ = λ(2x + 6) + μ

6x + 12 = (2λ)x + 6λ + μ

By comparing the coefficients of the power of x, we get

6 = 2λ ⇒ λ = 3

6λ + μ = 12

6(3) + μ = 12

μ = -6

So, l1 = ∫(3(2x + 6) – 6)/(x2 + 6x + 12) dx

=3∫ ((2x + 6))/(x2 + 6x + 12) dx – 6∫1/(x2 + 2x(3) + (3)2 – (3)2 + 12) dx

= 3∫ (2x + 6)/(x2 + 6x + 12) dx + 6∫1/((x + 3)2 + (√3)2) dx

As we know that, ∫1/(x2 + a2) dx = 1/2 tan-1⁡(x/a) + c

So, we get

I1 = 3log⁡|x2 + 6x + 12| + 6/√3 tan-1⁡((x + 3)/√3) + c2

I1 = 3log⁡|x2 + 6x + 12| + 2√3 tan-1⁡((x + 3)/√3) + c2

Now put the value of I1 in eq(i), we get

l = x – 3log⁡|x2 + 6x + 12| + 2√3 tan-1((x + 3)/√3) + c

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