Here we provide RD Sharma Class 12 Ex 19.29 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 19.29 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.
| Textbook | NCERT |
| Class | Class 12th |
| Subject | Maths |
| Chapter | 19 |
| Exercise | 19.32 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 19.29 Solutions Chapter 19 Indefinite Integrals
Evaluate the following integrals:
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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