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Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.32
  • Last Updated : 07 Apr, 2021

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/(t2 − 3)(t)dt

= 2∫dt/(t2 − 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/[(t2 − 3 − 2)/2](t)dt

= 2∫dt/(t2 − 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/(t2 − 3)(t)dt

= ∫(dx/√(x + 2)] + 2∫dt/(t2 − 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 

= ∫(x2 − 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/(t2 − 3)(t)dt

= ∫√(x + 2)dx − ∫dx/[√(x + 2)] + 2∫dt/(t2 − 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/[(t2 − 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/[(x2 + 1)√x]dx

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

So, the equation becomes,

= 2∫t/[(t4 + 1)(t)]dt

= 2∫dt/(t4 + 1)

= 2∫(t/t2)/(t2 + 1/t2)dt

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

= ∫(1 + 1/t2)/[(t − 1/t)2 + 2]dt − ∫(1 − 1/t2)/[(t + 1/t)2 − 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/(y2 +2) − ∫dz/(z2 − 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/[(x2 + 2x + 2)√(x + 1)]dx

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

So, the equation becomes,

= 2∫(t2 − 1)(t)/[(t4 + 1)(t)]dt

= 2∫(t2 − 1)/(t4 + 1)dt

= 2∫(1 − 1/t2)/[(t + 1/t)2 − 2]dt

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

So, the equation becomes,

= 2∫dy/(y2 − 2)

= (2/2√2) log |(y − √2)/(y + √2)| + c

= (1/√2) log |(t2 + 1 − √2t)/(t2 + 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)√(x2 + 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)2 + 1]]dt

= −∫dt/√(2t2 + 2t + 1)

= −(1/√2)∫dt/√(t2 + 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)√(x2 + x + 1)]dx

Solution:

We have,

∫1/[(x + 1)√(x2 + x + 1)]dx

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

So, the equation becomes,

= −∫(1/t2)/[(1/t)√(1/t2 + 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/[(x2 − 1)√(x2 + 1)]dx

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

So, the equation becomes,

= −∫(1/t2)/[(1/t2 − 1)√(1/t2 + 1)]dt

= −∫t/[(1 − t2)√(1 + t2)]dt

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

=> tdt = ydy

So, the equation becomes,

= ∫ydy/(y2 − 2)y

= ∫dy/(y2 − 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 + √(x2 + 1)]/[√2x − √(x2 + 1)]| + c

Question 11. ∫x/[(x2 + 4)√(x2 + 1)]dx

Solution:

We have,

∫x/[(x2 + 4)√(x2 + 1)]dx

Let x2 + 1 = t2, so we get, 2xdx = 2tdt

=> xdx = tdt

So, the equation becomes,

= ∫t/(t2 + 3)(t)dt

= ∫dt/(t2 + 3)

= (1/√3) tan−1(t/√3) + c

= (1/√3) tan−1[√(x2 + 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/t2 + 1)√(1 − 1/t2)]dt

= −∫t/[(t2 + 1)√(t2 − 1)]dt

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

=> tdt = ydy

So, the equation becomes,

= −∫y/[(y2 + 2)(y)]dy

= −∫1/(y2 + 2)dy

= −(1/√2) tan−1(y/√2) + c

= −(1/√2) tan−1(√(t2 − 1)/√2) + c

= −(1/√2) tan−1(√(1 − x2)/√2x) + c

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

Solution:

We have,

∫1/[(2x2 + 3)√(x2 − 4)]dx 

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

So, the equation becomes,

= −∫(1/t2)/[(2/t2 + 3)√(1/t2 − 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) + √(3x2 − 12)]/[√(11x) − √(3x2 − 12)]| + c

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

Solution:

We have,

∫x/[(x2 + 4)√(x2 + 9)]dx 

Let x2 + 9 = y2, so we have 2xdx = 2ydy

=> xdx = ydy

So, the equation becomes,

= ∫y/[(y2 − 5)y]dy

= ∫1/(y2 − 5)dy

= (1/2√5) log |(y − √5)/(y + √5)| + c

= (1/2√5) log |(√(x2 + 9) − √5)/(√(x2 + 9) + √5)| + c

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