# RD Sharma Class 12 Ex 19.32 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.32 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 19.32 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.32 Solutions Chapter 19 Indefinite Integrals

### Question 1. ∫1/[(x − 1)√(x + 2)]dx

Solution:

We have,

∫1/[(x − 1)√(x + 2)]dx

Let x + 2 = t2, so we get, xdx = 2tdt

So, the equation becomes,

= ∫2t/(t− 3)(t)dt

= 2∫dt/(t− 3)

= (2/2√3) log |(t − √3)/(t + √3)| + c

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

### Question 2. ∫1/[(x − 1)√(2x + 3)]dx

Solution:

We have,

∫1/[(x − 1)√(2x + 3)]dx

Let 2x + 3 = t2, so we have, 2dx = 2tdt,

=> dx = tdt

So, the equation becomes,

= ∫t/[(t− 3 − 2)/2](t)dt

= 2∫dt/(t− 5)

= (2/2√5) log |(t − √5)/(t + √5)| + c

= (1/√5) log |(√(2x + 3) − √5)/(√(2x + 3) + √5)| + c

### Question 3. ∫(x + 1)/[(x − 1)√(x + 2)]dx

Solution:

We have,

∫(x + 1)/[(x − 1)√(x + 2)]dx

= ∫(x − 1 + 2)/[(x − 1)√(x + 2)]dx

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

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

In second part, let x + 2 = t2, so we get, xdx = 2tdt

So, the equation becomes,

= ∫(dx/√(x + 2)] + ∫2t/(t− 3)(t)dt

= ∫(dx/√(x + 2)] + 2∫dt/(t− 3)

= 2√(x + 2) + c1 + (4/2√3) log |(t − √3)/(t + √3)| + c2

= 2√(x + 2) + (2/√3) log |(√(x − 2) − √3)/(√(x − 2)+√3)| + c

### Question 4. ∫x2/[(x − 1)√(x + 2)]dx

Solution:

We have,

∫x2/[(x − 1)√(x + 2)]dx

= ∫(x− 1 + 1)/[(x − 1)√(x + 2)]dx

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

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

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

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

In third part, let x + 2 = t2, so we get, xdx = 2tdt

So, the equation becomes,

= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + ∫2t/(t− 3)(t)dt

= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + 2∫dt/(t− 3)

= (2/3)(x + 2)3/2 + c1 − 2√(x + 2) + c2 + (2/2√3) log |(t − √3)/(t + √3)| + c3

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

### Question 5. ∫x/[(x − 3)√(x + 1)]dx

Solution:

We have,

∫x/[(x − 3)√(x + 1)]dx

= ∫[(x − 3) + 3]/[(x − 3)√(x + 1)]dx

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

For second part, let x + 1 = t2, so we get, dx = 2tdt.

So, the equation becomes,

= ∫dx/[√(x + 1)] + 3∫2tdt/[(t− 4)(t)]

= 2√(x + 1) + c1 + (3/2) log |(t − 2)/(t + 2)| + c2

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

### Question 6. ∫1/[(x2 + 1)√x]dx

Solution:

We have,

∫1/[(x+ 1)√x]dx

Let x = t2, so we have, dx = 2tdt

So, the equation becomes,

= 2∫t/[(t+ 1)(t)]dt

= 2∫dt/(t+ 1)

= 2∫(t/t2)/(t+ 1/t2)dt

= ∫[1 + 1/t− (1 − 1/t2)]/(t+ 1/t2)dt

= ∫(1 + 1/t2)/[(t − 1/t)+ 2]dt − ∫(1 − 1/t2)/[(t + 1/t)− 2]dt

Let t − 1/t = y, so we have, (1 + 1/t2)dt = dy

Let t + 1/t = z, so we have, (1 − 1/t2)dt = dz

So, the equation becomes,

= ∫dy/(y+2) − ∫dz/(z− 2)

= (1/√2) tan−1(y/√2) − (1/2√2) log |(z − √2)/(z + √2)| + c

= (1/√2) tan−1[(t2−1)/√2t] − (1/2√2) log |[x + 1 − √(2x)]/[x + 1 + √(2x)]| + c

= (1/√2) tan−1[(x−1)/√(2x)] − (1/2√2) log |[x + 1 − √(2x)]/[x + 1 + √(2x)]| + c

### Question 7. ∫x/[(x2 + 2x + 2)√(x + 1)]dx

Solution:

We have,

∫x/[(x+ 2x + 2)√(x + 1)]dx

Let x + 1 = t2, so we have, dx = 2tdt

So, the equation becomes,

= 2∫(t− 1)(t)/[(t+ 1)(t)]dt

= 2∫(t− 1)/(t+ 1)dt

= 2∫(1 − 1/t2)/[(t + 1/t)− 2]dt

Let t + 1/t = y, so we have, (1 − 1/t2)dt = dy

So, the equation becomes,

= 2∫dy/(y− 2)

= (2/2√2) log |(y − √2)/(y + √2)| + c

= (1/√2) log |(t+ 1 − √2t)/(t+ 1 + √2t)| + c

= (1/√2) log |[x + 2 − √(2x + 2)]/[x + 2 + √(2x + 2)]| + c

### Question 8. ∫1/[(x − 1)√(x2 + 1)]dx

Solution:

We have,

∫1/[(x − 1)√(x+ 1)]dx

Let x − 1 = 1/t, so we have dx = (−1/t2)dt

So, the equation becomes,

= −∫(1/t2)/[(1/t)√[(1 + 1/t)+ 1]]dt

= −∫dt/√(2t+ 2t + 1)

= −(1/√2)∫dt/√(t+ t + 1/2)

= −(1/√2)∫dt/√[(t + 1/2)+ 1/4]

= −(1/√2) log |(t + 1/2) + √[(t + 1/2)+ 1/4]| + c

= −(1/√2) log |(1/(x − 1) + 1/2) + √[(1/(x−1) + 1/2)+ 1/4]| + c

### Question 9. ∫1/[(x + 1)√(x2 + x + 1)]dx

Solution:

We have,

∫1/[(x + 1)√(x+ x + 1)]dx

Let x + 1 = 1/t, so we have dx = (−1/t2)dt

So, the equation becomes,

= −∫(1/t2)/[(1/t)√(1/t+ 1/t − 1)]dt

= −∫dt/√(1 + t − t2)

= −∫dt/√[5/4 − (1/4 − t + t2)]

= −∫dt/√[5/4 − (t − 1/2)2]

= − sin−1[(t − 1/2)/(√5/2)] + c

= − sin−1[(2t − 1)/√5] + c

= − sin−1[(1 − x)/[√5(x + 1)]] + c

### Question 10. ∫1/[(x2 − 1)√(x2 + 1)]dx

Solution:

We have,

∫1/[(x− 1)√(x+ 1)]dx

Let x = 1/t, so we get, dx = (−1/t2)dt

So, the equation becomes,

= −∫(1/t2)/[(1/t− 1)√(1/t+ 1)]dt

= −∫t/[(1 − t2)√(1 + t2)]dt

Let 1 + t2 = y2, so we have, 2tdt = 2ydy

=> tdt = ydy

So, the equation becomes,

= ∫ydy/(y− 2)y

= ∫dy/(y− 2)

= (1/2√2) log |(y − √2)/(y + √2)| + c

= (1/2√2) log |(y − √2)/(y + √2)| + c

= (1/2√2) log |[√(1 + t2) − √2]/[√(1 + t2) + √2]| + c

= −(1/2√2) log |[√2x + √(x+ 1)]/[√2x − √(x+ 1)]| + c

### Question 11. ∫x/[(x2 + 4)√(x2 + 1)]dx

Solution:

We have,

∫x/[(x+ 4)√(x+ 1)]dx

Let x+ 1 = t2, so we get, 2xdx = 2tdt

=> xdx = tdt

So, the equation becomes,

= ∫t/(t+ 3)(t)dt

= ∫dt/(t+ 3)

= (1/√3) tan−1(t/√3) + c

= (1/√3) tan−1[√(x+ 1)/√3] + c

### Question 12. ∫1/[(1 + x2)√(1 − x2)]dx

Solution:

We have,

∫1/[(1 + x2)√(1 − x2)]dx

Let x = 1/t, so we get, dx = (−1/t2)dt

So, the equation becomes,

= −∫(1/t2)/[(1/t+ 1)√(1 − 1/t2)]dt

= −∫t/[(t+ 1)√(t− 1)]dt

Let t− 1 = y2, so we get, 2tdt = 2ydy

=> tdt = ydy

So, the equation becomes,

= −∫y/[(y+ 2)(y)]dy

= −∫1/(y+ 2)dy

= −(1/√2) tan−1(y/√2) + c

= −(1/√2) tan−1(√(t− 1)/√2) + c

= −(1/√2) tan−1(√(1 − x2)/√2x) + c

### Question 13. ∫1/[(2x2 + 3)√(x2 − 4)]dx

Solution:

We have,

∫1/[(2x+ 3)√(x− 4)]dx

Let x = 1/t, so we have dx = (−1/t2)dt

So, the equation becomes,

= −∫(1/t2)/[(2/t+ 3)√(1/t− 4)]dt

= −∫t/[(2 + 3t2)√(1−4t2)]dt

Let 1 − 4t2 = y2, so we get, −8tdt = 2ydy

So, the equation becomes,

= (1/4) ∫y/[(11 − 3y2)y/4]dy

= (1/3) ∫1/(11/3 − y2)dy

= (1/2√33) log |[y − √(11/3)]/[y + √(11/3)]| + c

= (1/2√33) log |[√(1 − 4t2) − √(11/3)]/[√(1 + 4t2) + √(11/3)]| + c

= (1/2√33) log |[√(11x) + √(3x− 12)]/[√(11x) − √(3x− 12)]| + c

### Question 14. ∫x/[(x2 + 4)√(x2 + 9)]dx

Solution:

We have,

∫x/[(x+ 4)√(x+ 9)]dx

Let x+ 9 = y2, so we have 2xdx = 2ydy

=> xdx = ydy

So, the equation becomes,

= ∫y/[(y− 5)y]dy

= ∫1/(y− 5)dy

= (1/2√5) log |(y − √5)/(y + √5)| + c

= (1/2√5) log |(√(x+ 9) − √5)/(√(x+ 9) + √5)| + c

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