Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.29

• Last Updated : 20 Jan, 2022

Question 1. ∫(x + 1)√(x2 – x + 1)dx

Solution:

We have,

∫(x + 1)√(x2 – x + 1)dx

Let x + 1 = a d(x2 – x + 1)/dx + b

=> x + 1 = a(2x – 1) + b

On comparing both sides, we get,

=> 2a = 1 and b – a = 1

=> a = 1/2 and b = 1 + 1/2 = 3/2

So, our equation becomes,

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

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

For first part, let x2 – x + 1 = t, so we have, (2x – 1)dx = dt

So, we have,

= (1/2)∫√tdt + (3/2)∫√[(x – 1/2)2 + (√3/2)2]dx

= (1/2)(2/3)(t3/2) + (3/2)∫√[(x – 1/2)2 + (√3/2)2]dx

= (1/3)(x2 – x + 1)3/2 + (3/2)[(1/2)(x – 1/2)√(x2 – x + 1) + (3/8)log|(x – 1/2) + √(x2 – x + 1)|] + c

= (1/3)(x2 – x + 1)3/2 + (3/8)(2x – 1)√(x2 – x + 1) + (9/16) log|(x – 1/2) + √(x2 – x + 1)| + c

Question 2. ∫(x + 1)√(2x2 + 3)dx

Solution:

We have,

∫(x + 1)√(2x2 + 3)dx

Let x+1 = a d(2x2 + 3)/dx + b

=> x + 1 = a(4x) + b

On comparing both sides, we get,

=> 4a= 1 and b = 1

=> a = 1/4 and b = 1

So, the equation becomes,

= ∫[(1/4)(4x) + 1]√(2x2 + 3)dx

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

For first part, let 2x2 + 3 = t, so we have, 4xdx = dt

So, we have,

= (1/4)∫√tdt + √2∫√(x2+3/2)dx

= (1/4)(2/3)(t3/2) + √2[(x/2)√(x2 + 3/2) + (3/4) log |x + √(x2 + 3/2)|] + c

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

Question 3. ∫(2x – 5)√(2 + 3x – x2)dx

Solution:

We have,

∫(2x – 5)√(2 + 3x – x2)dx

Let 2x – 5 = a d(2 + 3x – x2)/dx + b

=> 2x – 5 = a(3 – 2x) + b

On comparing both sides, we get,

=> –2a= 2 and b + 3a = –5

=> a = –1 and b = –5 – 3(–1) = –2

So, the equation becomes,

= ∫[(–1)(3 – 2x) – 2]√(2 + 3x – x2)dx

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

For first part, let 2 + 3x – x2 = t, so we have, (3 – 2x)dx = dt

So, we have,

= –∫√tdt – 2∫√[(17/4) – (9/4 – 3x – x2)]dx

= –(2/3)(t3/2) – 2∫√[(√17/2)2 – (x – 3/2)2]dx

= –(2/3)(2 + 3x – x2)3/2 – 2[(1/2)(x – 3/2)√(2 + 3x – x2) + (17/8) sin-1[(x – 3/2)/(√17/2)]] + c

= –(2/3)(2 + 3x – x2)3/2 – (1/2)(2x – 3)√(2 + 3x – x2) – (17/8) sin-1[(2x – 3)/√17] + c

Question 4. ∫(x + 2)√(x2 + x + 1)dx

Solution:

We have,

∫(x + 2)√(x2 + x + 1)dx

Let x + 2 = a d(x2 + x + 1)/dx + b

=> x + 2 = a(2x + 1)+b

On comparing both sides, we get,

=> 2a = 1 and a + b = 2

=> a = 1/2 and b = 2 – 1/2 = 3/2

So, the equation becomes,

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

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

For first part, let x2 + x + 1 = t, so we have, (2x + 1)dx = dt

So, we have,

= (1/2)∫√tdt + (3/2)∫√[(x + 1/2)2 + (√3/2)2]dx

=  (1/2)(2/3)(t3/2) + (3/2)[(1/2)(x + 1/2)√(x2 + x + 1) + (3/8) log|(x + 1/2) + √(x2 + x + 1)|] + c

= (1/3)(x2 + x + 1)3/2 + (3/8)(2x + 1)√(x2 + x + 1) + (9/16) log|(x + 1/2) + √(x2 + x + 1)| + c

Question 5. ∫(4x + 1)√(x2 – x – 2)dx

Solution:

We have,

∫(4x + 1)√(x2 – x – 2)dx

Let 4x + 1 = a d(x2 – x – 2)/dx + b

=> 4x + 1 = a (2x – 1) + b

Comparing both sides, we get,

=> 2a = 4 and b – a = 1

=> a = 2 and b = 1 + 2

=> a = 2 and b = 3

So, the equation becomes,

= ∫[2(2x – 1) + 3]√(x2 – x – 2)dx

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

For first part, let x2 – x – 2 = t, so we have, (2x – 1)dx = dt

So, we have,

= 2∫√tdt + 3∫√(x2 – x – 2)dx

= 2∫√tdt + 3∫√[(x – 1/2)2 – (3/2)2]dx

= 2(2/3)(t3/2) + 3[(1/2)(x – 1/2)√(x2 – x – 2) – (9/8) log|(x – 1/2) + √(x2 – x – 2)|] + c

= (4/3)(x2 – x – 2)3/2 + (3/4)(2x – 1)√(x2 – x – 2) – (27/8) log|(x – 1/2) + √(x2 – x – 2)| + c

Question 6. ∫(x – 2)√(2x2 – 6x + 5)dx

Solution:

We have,

∫(x – 2)√(2x2 – 6x + 5)dx

Let x – 2 = a d(2x2 – 6x + 5)/dx + b

=> x – 2 = a(4x – 6) + b

On comparing both sides, we get,

=> 4a = 1 and b – 6a = –2

=> a = 1/4 and b = –2 + 6(1/4)

=> a = 1/4 and b = –1/2

So, the equation becomes,

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

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

For first part, let 2x2 – 6x + 5 = t, so we have, (4x – 6)dx = dt

So, we have,

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

= (1/4)(2/3)(t3/2) – (√2/2)∫√[(x – 3/2)2 + (1/2)2]dx

= (1/6)(2x2 – 6x + 5)3/2 – (1/√2)[(1/2)(x – 3/2)√(x2 – 3x + 5/2) + (1/8) log|(x – 3/2) + √(x2 – 3x + 5/2)|] + c

= (1/6)(2x2 – 6x + 5)3/2 – (1/8)(2x – 3)√(2x2 – 6x + 5) – (1/8√2) log|(x – 3/2) + √(x2 – 3x + 5/2)| + c

Question 7. ∫(x + 1)√(x2 + x + 1)dx

Solution:

We have,

∫(x + 1)√(x2 + x + 1)dx

Let x + 1 = a d(x2 + x + 1)/dx + b

=> x + 1 = a(2x + 1)+b

On comparing both sides, we get,

=> 2a = 1 and a + b = 1

=> a = 1/2 and b = 1/2

So, the equation becomes,

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

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

For first part, let x2 + x + 1 = t, so we have, (2x + 1)dx = dt

So we have,

= (1/2)∫√tdt + (1/2)∫√[(x + 1/2)2 + (√3/2)2]dx

= (1/2)(2/3)(t3/2) + (1/2)[(1/2)(x + 1/2)√(x2 + x + 1) + (3/8) log|(x + 1/2) + √(x2 + x + 1)|] + c

= (1/3)(x2 + x + 1)3/2 + (1/8)(2x + 1)√(x2 + x + 1) + (3/16)log|(x + 1/2) + √(x2 + x + 1)| + c

Question 8. ∫(2x + 3)√(x2 + 4x + 3)dx

Solution:

We have,

∫(2x + 3)√(x2 + 4x + 3)dx

Let 2x + 3 = a d(x2 + 4x + 3)/dx + b

=> 2x + 3 = a(2x + 4) + b

On comparing both sides, we get,

=> 2a = 2 and 4a + b = 3

=> a = 1 and b = 3 – 4 = –1

So, the equation becomes,

= ∫[2x + 4 + (–1)]√(x2 + 4x + 3)dx

= ∫(2x + 4)√(x2 + 4x + 3)dx – ∫√(x2 + 4x + 3)dx

For first part, let x2 + 4x + 3 = t, so we have, (2x + 4)dx = dt

So, we have,

= ∫√tdt – ∫√(x2 + 4x + 3)dx

= (2/3)(t3/2) – ∫√[(x + 2)2–1]dx

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

Question 9. ∫(2x – 5)√(x2 – 4x + 3)dx

Solution:

We have,

∫(2x – 5)√(x2 – 4x + 3)dx

Let 2x – 5 = a d(x2 – 4x + 3)/dx + b

=> 2x – 5 = a(2x – 4) + b

On comparing both sides, we get,

=> 2a = 2 and b – 4a = –5

=> a = 1 and b = –5 + 4(1) = –1

So, the equation becomes,

= ∫2x – 4 + (–1)]√(x2 – 4x + 3)dx

= ∫(2x + 4)√(x2 – 4x + 3)dx – ∫√(x2 – 4x + 3)dx

For first part, let x2 – 4x + 3 = t, so we have, (2x – 4)dx = dt

So, we have,

= ∫√tdt – ∫√(x2 – 4x + 3)dx

= (2/3)(t3/2) – ∫√[(x – 2)2 – 1]dx

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

Question 10. ∫x√(x2 + x)dx

Solution:

We have,

∫x√(x2 + x)dx

Let x = a d(x2 + x)/dx + b

=> x = a(2x + 1) + b

On comparing both sides, we get,

=> 2a = 1 and a + b = 0

=> a = 1/2 and b = –1/2

So, the equation becomes,

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

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

For first part, let x2 + x = t, so we have, (2x + 1)dx = dt

So, we have,

= (1/2)∫√tdt – 1/2∫√(x2 + x)dx

= (1/2)(2/3)(t3/2) – (1/2)∫√[(x + 1/2)2 – (1/2)2]dx

= (1/3)(t3/2) – 1/2[(1/2)(x + 1/2)√(x2 + x)] + (1/8)log|(x + 1/2) + √(x2 + x)|] + c

= (1/3)(x2 + x)3/2 – (1/8)(2x + 1)√(x2 + x) + (1/16)log|(x + 1/2) + √(x2 + x)| + c

Question 11. ∫(x – 3)√(x2 + 3x – 18)dx

Solution:

We have,

∫(x – 3)√(x2 + 3x – 18)dx

Let x – 3 = a d(x2 + 3x – 18)/dx + b

=> x – 3 = a(2x + 3) + b

On comparing both sides, we get,

=> 2a = 1 and 3a + b = –3

=> a = 1/2 and b = –3 – 3/2 = –9/2

So, the equation becomes,

= ∫[(1/2)(2x + 3) – (9/2)]√(x2 + 3x – 18)dx

= (1/2)∫(2x + 3)√(x2 + 3x – 18)dx – (9/2)∫√(x2 + 3x – 18)dx

For first part, let x2 + 3x – 18 = t, so we have, (2x + 3)dx = dt

So, we have,

= (1/2)∫√tdt – (9/2)∫√(x2 + 3x – 18)dx

= (1/2)(2/3)(t3/2) – (9/2)∫√[(x + 3/2)2 – (9/2)2])dx

= (1/3)(x2 + 3x – 18)3/2 – (9/2)[(x + 3/2)√(x2 + 3x – 18) – (81/8) log |(x + 3/2) + √(x2 + 3x – 18)|] + c

= (1/3)(x2 + 3x – 18)3/2 – (9/8)(2x + 3)√(x2 + 3x – 18) + (729/16) log |(x + 3/2) + √(x2 + 3x – 18)| + c

Question 12. ∫(x + 3)√(3 – 4x – x2)dx

Solution:

We have,

∫(x + 3)√(3 – 4x – x2)dx

Let x + 3 = a d(3 – 4x – x2)/dx + b

=> x + 3 = a(–4 – 2x) + b

On comparing both sides, we get,

=> –2a = 1 and b – 4a = 3

=> a = –1/2 and b = 3 + 4(–1/2) = 1

So, the equation becomes,

= ∫[(–1/2)(–4 – 2x) + 1]√(3 – 4x – x2)dx

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

For first part, let 3 – 4x – x2 = t, so we have, (–4 – 2x)dx = dt

So, we have,

= (–1/2)∫√tdt + ∫√(3 – 4x – x2)dx

= (–1/2)(2/3)(t3/2) + ∫√[(√7)2 – (x + 2)2]dx

= (–1/3)(3 – 4x – x2)3/2 + (1/2)[(x + 2)√(3 – 4x – x2) + 7 tan–1[(x + 2)/√7]] + c

= (–1/3)(3 – 4x – x2)3/2 + (1/2)(x + 2)√(3 – 4x – x2) + (7/2) tan–1[(x + 2)/√7] + c

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