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

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 + c₁ 

₌ 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) + c₂ 

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