RD Sharma Class 12 Ex 19.20 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.20 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.20 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

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

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