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

### Evaluate the following integrals:

### Question 1. ∫(x + 1)√(x^{2 }– x + 1)dx

**Solution:**

We have,

∫(x + 1)√(x

^{2 }– x + 1)dxLet x + 1 = a d(x

^{2 }– 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]√(x

^{2 }– x + 1)dx= (1/2)∫(2x – 1)√(x

^{2 }– x + 1)dx + (3/2)∫√(x^{2 }– x + 1)dxFor first part, let x

^{2 }– x + 1 = t, so we have, (2x – 1)dx = dtSo, we have,

= (1/2)∫√tdt + (3/2)∫√[(x – 1/2)

^{2 }+ (√3/2)^{2}]dx= (1/2)(2/3)(t

^{3/2}) + (3/2)∫√[(x – 1/2)^{2 }+ (√3/2)^{2}]dx= (1/3)(x

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

^{2 }– x + 1)^{3/2}+ (3/8)(2x – 1)√(x^{2 }– x + 1) + (9/16) log|(x – 1/2) + √(x^{2 }– x + 1)| + c

### Question 2. ∫(x + 1)√(2x^{2 }+ 3)dx

**Solution:**

We have,

∫(x + 1)√(2x

^{2 }+ 3)dxLet x+1 = a d(2x

^{2 }+ 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]√(2x

^{2 }+ 3)dx= ∫(1/4)(4x)(x + 1)√(2x

^{2 }+ 3)dx + ∫√(2x^{2 }+ 3)dxFor first part, let 2x

^{2 }+ 3 = t, so we have, 4xdx = dtSo, we have,

= (1/4)∫√tdt + √2∫√(x

^{2}+3/2)dx= (1/4)(2/3)(t

^{3/2}) + √2[(x/2)√(x^{2 }+ 3/2) + (3/4) log |x + √(x^{2 }+ 3/2)|] + c= (1/6)(2x

^{2 }+ 3)^{3/2}+ (x/2)√(2x^{2 }+ 3) + (3/2√2) log |x + √(x^{2 }+ 3/2)| + c

### Question 3. ∫(2x – 5)√(2 + 3x – x^{2})dx

**Solution:**

We have,

∫(2x – 5)√(2 + 3x – x

^{2})dxLet 2x – 5 = a d(2 + 3x – x

^{2})/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 – x

^{2})dx= –∫(3 – 2x)√(2 + 3x – x

^{2})dx – 2∫√(2 + 3x – x^{2})dxFor first part, let 2 + 3x – x

^{2}= t, so we have, (3 – 2x)dx = dtSo, we have,

= –∫√tdt – 2∫√[(17/4) – (9/4 – 3x – x

^{2})]dx= –(2/3)(t

^{3/2}) – 2∫√[(√17/2)^{2 }– (x – 3/2)^{2}]dx= –(2/3)(2 + 3x – x

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

^{2})^{3/2}– (1/2)(2x – 3)√(2 + 3x – x^{2}) – (17/8) sin^{-1}[(2x – 3)/√17] + c

### Question 4. ∫(x + 2)√(x^{2 }+ x + 1)dx

**Solution:**

We have,

∫(x + 2)√(x

^{2 }+ x + 1)dxLet x + 2 = a d(x

^{2 }+ 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]√(x

^{2 }+ x + 1)dx= (1/2)∫(2x + 1)√(x

^{2 }+ x + 1)dx + (3/2)∫√(x^{2 }+ x + 1)dxFor first part, let x

^{2 }+ x + 1 = t, so we have, (2x + 1)dx = dtSo, we have,

= (1/2)∫√tdt + (3/2)∫√[(x + 1/2)

^{2 }+ (√3/2)^{2}]dx= (1/2)(2/3)(t

^{3/2}) + (3/2)[(1/2)(x + 1/2)√(x^{2 }+ x + 1) + (3/8) log|(x + 1/2) + √(x^{2 }+ x + 1)|] + c= (1/3)(x

^{2 }+ x + 1)^{3/2}+ (3/8)(2x + 1)√(x^{2 }+ x + 1) + (9/16) log|(x + 1/2) + √(x^{2 }+ x + 1)| + c

### Question 5. ∫(4x + 1)√(x^{2 }– x – 2)dx

**Solution:**

We have,

∫(4x + 1)√(x

^{2 }– x – 2)dxLet 4x + 1 = a d(x

^{2 }– 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]√(x

^{2 }– x – 2)dx= 2∫(2x – 1)√(x

^{2 }– x – 2)dx + 3∫√(x^{2 }– x – 2)dxFor first part, let x

^{2 }– x – 2 = t, so we have, (2x – 1)dx = dtSo, we have,

= 2∫√tdt + 3∫√(x

^{2 }– x – 2)dx= 2∫√tdt + 3∫√[(x – 1/2)

^{2 }– (3/2)^{2}]dx= 2(2/3)(t

^{3/2}) + 3[(1/2)(x – 1/2)√(x^{2 }– x – 2) – (9/8) log|(x – 1/2) + √(x^{2 }– x – 2)|] + c= (4/3)(x

^{2 }– x – 2)^{3/2 }+ (3/4)(2x – 1)√(x^{2 }– x – 2) – (27/8) log|(x – 1/2) + √(x^{2 }– x – 2)| + c

### Question 6. ∫(x – 2)√(2x^{2 }– 6x + 5)dx

**Solution:**

We have,

∫(x – 2)√(2x

^{2 }– 6x + 5)dxLet x – 2 = a d(2x

^{2 }– 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)]√(2x

^{2 }– 6x + 5)dx= (1/4)∫(4x – 6)√(2x

^{2 }– 6x + 5)dx – (1/2)∫√(2x^{2 }– 6x + 5)dxFor first part, let 2x

^{2 }– 6x + 5 = t, so we have, (4x – 6)dx = dtSo, we have,

= (1/4)∫√tdt – (√2/2)∫√(x

^{2 }– 3x + 5/2)dx= (1/4)(2/3)(t

^{3/2}) – (√2/2)∫√[(x – 3/2)^{2 }+ (1/2)^{2}]dx= (1/6)(2x

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

^{2 }– 6x + 5)^{3/2}– (1/8)(2x – 3)√(2x^{2 }– 6x + 5) – (1/8√2) log|(x – 3/2) + √(x^{2 }– 3x + 5/2)| + c

### Question 7. ∫(x + 1)√(x^{2 }+ x + 1)dx

**Solution:**

We have,

∫(x + 1)√(x

^{2 }+ x + 1)dxLet x + 1 = a d(x

^{2 }+ 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]√(x

^{2 }+ x + 1)dx= (1/2)∫(2x + 1)√(x

^{2 }+ x + 1)dx + (1/2)∫√(x^{2 }+ x + 1)dxFor first part, let x

^{2 }+ x + 1 = t, so we have, (2x + 1)dx = dtSo we have,

= (1/2)∫√tdt + (1/2)∫√[(x + 1/2)

^{2 }+ (√3/2)^{2}]dx= (1/2)(2/3)(t

^{3/2}) + (1/2)[(1/2)(x + 1/2)√(x^{2 }+ x + 1) + (3/8) log|(x + 1/2) + √(x^{2 }+ x + 1)|] + c= (1/3)(x

^{2 }+ x + 1)^{3/2}+ (1/8)(2x + 1)√(x^{2 }+ x + 1) + (3/16)log|(x + 1/2) + √(x^{2 }+ x + 1)| + c

### Question 8. ∫(2x + 3)√(x^{2 }+ 4x + 3)dx

**Solution: **

We have,

∫(2x + 3)√(x

^{2 }+ 4x + 3)dxLet 2x + 3 = a d(x

^{2 }+ 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)]√(x

^{2 }+ 4x + 3)dx= ∫(2x + 4)√(x

^{2 }+ 4x + 3)dx – ∫√(x^{2 }+ 4x + 3)dxFor first part, let x

^{2 }+ 4x + 3 = t, so we have, (2x + 4)dx = dtSo, we have,

= ∫√tdt – ∫√(x

^{2 }+ 4x + 3)dx= (2/3)(t

^{3/2}) – ∫√[(x + 2)^{2}–1]dx= (2/3)(x

^{2 }+ 4x + 3)^{3/2}– (1/2)(x + 2)√(x^{2 }+ 4x + 3) + (1/2)log |x + 2 + √(x^{2 }+ 4x + 3)| + c

### Question 9. ∫(2x – 5)√(x^{2 }– 4x + 3)dx

**Solution:**

We have,

∫(2x – 5)√(x

^{2 }– 4x + 3)dxLet 2x – 5 = a d(x

^{2 }– 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)]√(x

^{2 }– 4x + 3)dx= ∫(2x + 4)√(x

^{2 }– 4x + 3)dx – ∫√(x^{2 }– 4x + 3)dxFor first part, let x

^{2 }– 4x + 3 = t, so we have, (2x – 4)dx = dtSo, we have,

= ∫√tdt – ∫√(x

^{2 }– 4x + 3)dx= (2/3)(t

^{3/2})^{ }– ∫√[(x – 2)^{2 }– 1]dx= (2/3)(x

^{2 }– 4x + 3)^{3/2}– (1/2)(x – 2)√(x^{2 }– 4x + 3) + (1/2)log |(x – 2) + √(x^{2 }– 4x + 3)| + c

### Question 10. ∫x√(x^{2 }+ x)dx

**Solution:**

We have,

∫x√(x

^{2 }+ x)dxLet x = a d(x

^{2 }+ 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)]√(x

^{2 }+ x)dx= (1/2)∫(2x + 1)√(x

^{2 }+ x)dx – 1/2∫√(x^{2 }+ x)dxFor first part, let x

^{2 }+ x = t, so we have, (2x + 1)dx = dtSo, we have,

= (1/2)∫√tdt – 1/2∫√(x

^{2 }+ x)dx= (1/2)(2/3)(t

^{3/2})^{ }– (1/2)∫√[(x + 1/2)^{2 }– (1/2)^{2}]dx= (1/3)(t

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

^{2 }+ x)^{3/2}– (1/8)(2x + 1)√(x^{2 }+ x) + (1/16)log|(x + 1/2) + √(x^{2 }+ x)| + c

### Question 11. ∫(x – 3)√(x^{2 }+ 3x – 18)dx

**Solution:**

We have,

∫(x – 3)√(x

^{2 }+ 3x – 18)dxLet x – 3 = a d(x

^{2 }+ 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)]√(x

^{2 }+ 3x – 18)dx= (1/2)∫(2x + 3)√(x

^{2 }+ 3x – 18)dx – (9/2)∫√(x^{2 }+ 3x – 18)dxFor first part, let x

^{2 }+ 3x – 18 = t, so we have, (2x + 3)dx = dtSo, we have,

= (1/2)∫√tdt – (9/2)∫√(x

^{2 }+ 3x – 18)dx= (1/2)(2/3)(t

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

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

^{2 }+ 3x – 18)^{3/2}– (9/8)(2x + 3)√(x^{2 }+ 3x – 18) + (729/16) log |(x + 3/2) + √(x^{2 }+ 3x – 18)| + c

### Question 12. ∫(x + 3)√(3 – 4x – x^{2})dx

**Solution:**

We have,

∫(x + 3)√(3 – 4x – x

^{2})dxLet x + 3 = a d(3 – 4x – x

^{2})/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 – x

^{2})dx= (–1/2)∫(–4 – 2x)√(3 – 4x – x

^{2})dx + ∫√(3 – 4x – x^{2})dxFor first part, let 3 – 4x – x

^{2}= t, so we have, (–4 – 2x)dx = dtSo, we have,

= (–1/2)∫√tdt + ∫√(3 – 4x – x

^{2})dx= (–1/2)(2/3)(t

^{3/2}) + ∫√[(√7)^{2 }– (x + 2)^{2}]dx= (–1/3)(3 – 4x – x

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

^{2})^{3/2}+ (1/2)(x + 2)√(3 – 4x – x^{2}) + (7/2) tan^{–1}[(x + 2)/√7] + c