# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.32

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

**Solution:**

We have,

∫1/[(x − 1)√(x + 2)]dx

Let x + 2 = t

^{2}, so we get, xdx = 2tdtSo, the equation becomes,

= ∫2t/(t

^{2 }− 3)(t)dt= 2∫dt/(t

^{2 }− 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 = t

^{2}, so we have, 2dx = 2tdt,=> dx = tdt

So, the equation becomes,

= ∫t/[(t

^{2 }− 3 − 2)/2](t)dt= 2∫dt/(t

^{2 }− 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 = t

^{2}, so we get, xdx = 2tdtSo, the equation becomes,

= ∫(dx/√(x + 2)] + ∫2t/(t

^{2 }− 3)(t)dt= ∫(dx/√(x + 2)] + 2∫dt/(t

^{2 }− 3)= 2√(x + 2) + c

_{1}+ (4/2√3) log |(t − √3)/(t + √3)| + c_{2}= 2√(x + 2) + (2/√3) log |(√(x − 2) − √3)/(√(x − 2)+√3)| + c

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

**Solution:**

We have,

∫x

^{2}/[(x − 1)√(x + 2)]dx= ∫(x

^{2 }− 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 = t

^{2}, so we get, xdx = 2tdtSo, the equation becomes,

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

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

^{2 }− 3)= (2/3)(x + 2)

^{3/2}+ c_{1}− 2√(x + 2) + c_{2}+ (2/2√3) log |(t − √3)/(t + √3)| + c_{3}= (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 = t

^{2}, so we get, dx = 2tdt.So, the equation becomes,

= ∫dx/[√(x + 1)] + 3∫2tdt/[(t

^{2 }− 4)(t)]= 2√(x + 1) + c

_{1}+ (3/2) log |(t − 2)/(t + 2)| + c_{2}= 2√(x + 1) + (3/2) log |(√(x + 1) − 2)/(√(x + 1) + 2)| + c

### Question 6. ∫1/[(x^{2 }+ 1)√x]dx

**Solution:**

We have,

∫1/[(x

^{2 }+ 1)√x]dxLet x = t

^{2}, so we have, dx = 2tdtSo, the equation becomes,

= 2∫t/[(t

^{4 }+ 1)(t)]dt= 2∫dt/(t

^{4 }+ 1)= 2∫(t/t

^{2})/(t^{2 }+ 1/t^{2})dt= ∫[1 + 1/t

^{2 }− (1 − 1/t^{2})]/(t^{2 }+ 1/t^{2})dt= ∫(1 + 1/t

^{2})/[(t − 1/t)^{2 }+ 2]dt − ∫(1 − 1/t^{2})/[(t + 1/t)^{2 }− 2]dtLet t − 1/t = y, so we have, (1 + 1/t

^{2})dt = dyLet t + 1/t = z, so we have, (1 − 1/t

^{2})dt = dzSo, the equation becomes,

= ∫dy/(y

^{2 }+2) − ∫dz/(z^{2 }− 2)= (1/√2) tan

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

^{−1}[(t^{2}−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/[(x^{2 }+ 2x + 2)√(x + 1)]dx

**Solution:**

We have,

∫x/[(x

^{2 }+ 2x + 2)√(x + 1)]dxLet x + 1 = t

^{2}, so we have, dx = 2tdtSo, the equation becomes,

= 2∫(t

^{2 }− 1)(t)/[(t^{4 }+ 1)(t)]dt= 2∫(t

^{2 }− 1)/(t^{4 }+ 1)dt= 2∫(1 − 1/t

^{2})/[(t + 1/t)^{2 }− 2]dtLet t + 1/t = y, so we have, (1 − 1/t

^{2})dt = dySo, the equation becomes,

= 2∫dy/(y

^{2 }− 2)= (2/2√2) log |(y − √2)/(y + √2)| + c

= (1/√2) log |(t

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

### Question 8. ∫1/[(x − 1)√(x^{2 }+ 1)]dx

**Solution:**

We have,

∫1/[(x − 1)√(x

^{2 }+ 1)]dxLet x − 1 = 1/t, so we have dx = (−1/t

^{2})dtSo, the equation becomes,

= −∫(1/t

^{2})/[(1/t)√[(1 + 1/t)^{2 }+ 1]]dt= −∫dt/√(2t

^{2 }+ 2t + 1)= −(1/√2)∫dt/√(t

^{2 }+ t + 1/2)= −(1/√2)∫dt/√[(t + 1/2)

^{2 }+ 1/4]= −(1/√2) log |(t + 1/2) + √[(t + 1/2)

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

^{2 }+ 1/4]| + c

### Question 9. ∫1/[(x + 1)√(x^{2 }+ x + 1)]dx

**Solution:**

We have,

∫1/[(x + 1)√(x

^{2 }+ x + 1)]dxLet x + 1 = 1/t, so we have dx = (−1/t

^{2})dtSo, the equation becomes,

= −∫(1/t

^{2})/[(1/t)√(1/t^{2 }+ 1/t − 1)]dt= −∫dt/√(1 + t − t

^{2})= −∫dt/√[5/4 − (1/4 − t + t

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

**Solution:**

We have,

∫1/[(x

^{2 }− 1)√(x^{2 }+ 1)]dxLet x = 1/t, so we get, dx = (−1/t

^{2})dtSo, the equation becomes,

= −∫(1/t

^{2})/[(1/t^{2 }− 1)√(1/t^{2 }+ 1)]dt= −∫t/[(1 − t

^{2})√(1 + t^{2})]dtLet 1 + t

^{2}= y^{2}, so we have, 2tdt = 2ydy=> tdt = ydy

So, the equation becomes,

= ∫ydy/(y

^{2 }− 2)y= ∫dy/(y

^{2 }− 2)= (1/2√2) log |(y − √2)/(y + √2)| + c

= (1/2√2) log |(y − √2)/(y + √2)| + c

= (1/2√2) log |[√(1 + t

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

^{2 }+ 1)]/[√2x − √(x^{2 }+ 1)]| + c

### Question 11. ∫x/[(x^{2 }+ 4)√(x^{2 }+ 1)]dx

**Solution:**

We have,

∫x/[(x

^{2 }+ 4)√(x^{2 }+ 1)]dxLet x

^{2 }+ 1 = t^{2}, so we get, 2xdx = 2tdt=> xdx = tdt

So, the equation becomes,

= ∫t/(t

^{2 }+ 3)(t)dt= ∫dt/(t

^{2 }+ 3)= (1/√3) tan

^{−1}(t/√3) + c= (1/√3) tan

^{−1}[√(x^{2 }+ 1)/√3] + c

### Question 12. ∫1/[(1 + x^{2})√(1 − x^{2})]dx

**Solution:**

We have,

∫1/[(1 + x

^{2})√(1 − x^{2})]dxLet x = 1/t, so we get, dx = (−1/t

^{2})dtSo, the equation becomes,

= −∫(1/t

^{2})/[(1/t^{2 }+ 1)√(1 − 1/t^{2})]dt= −∫t/[(t

^{2 }+ 1)√(t^{2 }− 1)]dtLet t

^{2 }− 1 = y^{2}, so we get, 2tdt = 2ydy=> tdt = ydy

So, the equation becomes,

= −∫y/[(y

^{2 }+ 2)(y)]dy= −∫1/(y

^{2 }+ 2)dy= −(1/√2) tan

^{−1}(y/√2) + c= −(1/√2) tan

^{−1}(√(t^{2 }− 1)/√2) + c= −(1/√2) tan

^{−1}(√(1 − x^{2})/√2x) + c

### Question 13. ∫1/[(2x^{2 }+ 3)√(x^{2 }− 4)]dx

**Solution:**

We have,

∫1/[(2x

^{2 }+ 3)√(x^{2 }− 4)]dxLet x = 1/t, so we have dx = (−1/t

^{2})dtSo, the equation becomes,

= −∫(1/t

^{2})/[(2/t^{2 }+ 3)√(1/t^{2 }− 4)]dt= −∫t/[(2 + 3t

^{2})√(1−4t^{2})]dtLet 1 − 4t

^{2}= y^{2}, so we get, −8tdt = 2ydySo, the equation becomes,

= (1/4) ∫y/[(11 − 3y

^{2})y/4]dy= (1/3) ∫1/(11/3 − y

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

= (1/2√33) log |[√(1 − 4t

^{2}) − √(11/3)]/[√(1 + 4t^{2}) + √(11/3)]| + c= (1/2√33) log |[√(11x) + √(3x

^{2 }− 12)]/[√(11x) − √(3x^{2 }− 12)]| + c

### Question 14. ∫x/[(x^{2 }+ 4)√(x^{2 }+ 9)]dx

**Solution:**

We have,

∫x/[(x

^{2 }+ 4)√(x^{2 }+ 9)]dxLet x

^{2 }+ 9 = y^{2}, so we have 2xdx = 2ydy=> xdx = ydy

So, the equation becomes,

= ∫y/[(y

^{2 }− 5)y]dy= ∫1/(y

^{2 }− 5)dy= (1/2√5) log |(y − √5)/(y + √5)| + c

= (1/2√5) log |(√(x

^{2 }+ 9) − √5)/(√(x^{2 }+ 9) + √5)| + c

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