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| + 2 × \frac{1}{2(\frac{1}{2})} log⁡|\frac{(x-\frac{1}{2}-\frac{1}{2})}{(x-\frac{1}{2}+\frac{1}{2})}|+c_1        

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, \frac{(x^2 + 1)}{(x^2 - 5x + 6)} = 1 - \frac{5}{(x-2)} + \frac{10}{(x - 3)}

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

Summary

Indefinite Integration: The process of finding a function F(x) whose derivative is the given function f(x).

General Form: ∫ f(x) dx = F(x) + C, where F'(x) = f(x) and C is the constant of integration.

Basic Integration Rules:

- ∫ xⁿ dx = (xⁿ⁺¹)/(n+1) + C, for n ≠ -1

- ∫ 1/x dx = ln|x| + C

- ∫ e^x dx = e^x + C

- ∫ sin x dx = - cos x + C

- ∫ cos x dx = sin x + C

Integration Techniques:

- Substitution Method

- Integration by Parts

- Partial Fraction Decomposition

Trigonometric Integrals: Special techniques for integrating products and powers of trigonometric functions.