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

Question 1. âˆ«1/[(x âˆ’ 1)âˆš(x + 2)]dx

Solution:

We have,

âˆ«1/[(x âˆ’ 1)âˆš(x + 2)]dx

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

So, the equation becomes,

= âˆ«2t/(t2 âˆ’ 3)(t)dt

= 2âˆ«dt/(t2 âˆ’ 3)

= (2/2âˆš3) log |(t âˆ’ âˆš3)/(t + âˆš3)| + c

= (1/âˆš3) log |(âˆš(x âˆ’ 2) âˆ’ âˆš3)/(âˆš(x âˆ’ 2) + âˆš3)| + c

Question 2. âˆ«1/[(x âˆ’ 1)âˆš(2x + 3)]dx

Solution:

We have,

âˆ«1/[(x âˆ’ 1)âˆš(2x + 3)]dx

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

=> dx = tdt

So, the equation becomes,

= âˆ«t/[(t2 âˆ’ 3 âˆ’ 2)/2](t)dt

= 2âˆ«dt/(t2 âˆ’ 5)

= (2/2âˆš5) log |(t âˆ’ âˆš5)/(t + âˆš5)| + c

= (1/âˆš5) log |(âˆš(2x + 3) âˆ’ âˆš5)/(âˆš(2x + 3) + âˆš5)| + c

Question 3. âˆ«(x + 1)/[(x âˆ’ 1)âˆš(x + 2)]dx

Solution:

We have,

âˆ«(x + 1)/[(x âˆ’ 1)âˆš(x + 2)]dx

= âˆ«(x âˆ’ 1 + 2)/[(x âˆ’ 1)âˆš(x + 2)]dx

= âˆ«(x âˆ’ 1)/[(x âˆ’ 1)âˆš(x + 2)]dx + âˆ«2/[(x âˆ’ 1)âˆš(x + 2)]dx

= âˆ«(dx/âˆš(x + 2)] + 2âˆ«dx/[(x âˆ’ 1)âˆš(x + 2)]

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

So, the equation becomes,

= âˆ«(dx/âˆš(x + 2)] + âˆ«2t/(t2 âˆ’ 3)(t)dt

= âˆ«(dx/âˆš(x + 2)] + 2âˆ«dt/(t2 âˆ’ 3)

= 2âˆš(x + 2) + c1 + (4/2âˆš3) log |(t âˆ’ âˆš3)/(t + âˆš3)| + c2

= 2âˆš(x + 2) + (2/âˆš3) log |(âˆš(x âˆ’ 2) âˆ’ âˆš3)/(âˆš(x âˆ’ 2)+âˆš3)| + c

Question 4. âˆ«x2/[(x âˆ’ 1)âˆš(x + 2)]dx

Solution:

We have,

âˆ«x2/[(x âˆ’ 1)âˆš(x + 2)]dx

= âˆ«(x2 âˆ’ 1 + 1)/[(x âˆ’ 1)âˆš(x + 2)]dx

= âˆ«(x âˆ’ 1)(x + 1)/[(x âˆ’ 1)âˆš(x + 2)]dx + âˆ«dx/[(x âˆ’ 1)âˆš(x + 2)]

= âˆ«(x + 1)/[âˆš(x + 2)]dx + âˆ«dx/[(x âˆ’ 1)âˆš(x + 2)]

= âˆ«[(x + 2) âˆ’ 1]/[âˆš(x + 2)]dx + âˆ«dx/[(x âˆ’ 1)âˆš(x + 2)]

= âˆ«âˆš(x + 2)dx âˆ’ âˆ«dx/[âˆš(x + 2)] + âˆ«dx/[(x âˆ’ 1)âˆš(x + 2)]

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

So, the equation becomes,

= âˆ«âˆš(x + 2)dx âˆ’ âˆ«dx/[âˆš(x + 2)] + âˆ«2t/(t2 âˆ’ 3)(t)dt

= âˆ«âˆš(x + 2)dx âˆ’ âˆ«dx/[âˆš(x + 2)] + 2âˆ«dt/(t2 âˆ’ 3)

= (2/3)(x + 2)3/2 + c1 âˆ’ 2âˆš(x + 2) + c2 + (2/2âˆš3) log |(t âˆ’ âˆš3)/(t + âˆš3)| + c3

= (2/3)(x + 2)3/2 âˆ’ 2âˆš(x + 2) + (1/âˆš3) log |(âˆš(x âˆ’ 2) âˆ’ âˆš3)/(âˆš(x âˆ’ 2) + âˆš3)| + c

Question 5. âˆ«x/[(x âˆ’ 3)âˆš(x + 1)]dx

Solution:

We have,

âˆ«x/[(x âˆ’ 3)âˆš(x + 1)]dx

= âˆ«[(x âˆ’ 3) + 3]/[(x âˆ’ 3)âˆš(x + 1)]dx

= âˆ«dx/[âˆš(x + 1)] + 3âˆ«dx/[(x âˆ’ 3)âˆš(x + 1)]

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

So, the equation becomes,

= âˆ«dx/[âˆš(x + 1)] + 3âˆ«2tdt/[(t2 âˆ’ 4)(t)]

= 2âˆš(x + 1) + c1 + (3/2) log |(t âˆ’ 2)/(t + 2)| + c2

= 2âˆš(x + 1) + (3/2) log |(âˆš(x + 1) âˆ’ 2)/(âˆš(x + 1) + 2)| + c

Question 6. âˆ«1/[(x2 + 1)âˆšx]dx

Solution:

We have,

âˆ«1/[(x2 + 1)âˆšx]dx

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

So, the equation becomes,

= 2âˆ«t/[(t4 + 1)(t)]dt

= 2âˆ«dt/(t4 + 1)

= 2âˆ«(t/t2)/(t2 + 1/t2)dt

= âˆ«[1 + 1/t2 âˆ’ (1 âˆ’ 1/t2)]/(t2 + 1/t2)dt

= âˆ«(1 + 1/t2)/[(t âˆ’ 1/t)2 + 2]dt âˆ’ âˆ«(1 âˆ’ 1/t2)/[(t + 1/t)2 âˆ’ 2]dt

Let t âˆ’ 1/t = y, so we have, (1 + 1/t2)dt = dy

Let t + 1/t = z, so we have, (1 âˆ’ 1/t2)dt = dz

So, the equation becomes,

= âˆ«dy/(y2 +2) âˆ’ âˆ«dz/(z2 âˆ’ 2)

= (1/âˆš2) tanâˆ’1(y/âˆš2) âˆ’ (1/2âˆš2) log |(z âˆ’ âˆš2)/(z + âˆš2)| + c

= (1/âˆš2) tanâˆ’1[(t2âˆ’1)/âˆš2t] âˆ’ (1/2âˆš2) log |[x + 1 âˆ’ âˆš(2x)]/[x + 1 + âˆš(2x)]| + c

= (1/âˆš2) tanâˆ’1[(xâˆ’1)/âˆš(2x)] âˆ’ (1/2âˆš2) log |[x + 1 âˆ’ âˆš(2x)]/[x + 1 + âˆš(2x)]| + c

Question 7. âˆ«x/[(x2 + 2x + 2)âˆš(x + 1)]dx

Solution:

We have,

âˆ«x/[(x2 + 2x + 2)âˆš(x + 1)]dx

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

So, the equation becomes,

= 2âˆ«(t2 âˆ’ 1)(t)/[(t4 + 1)(t)]dt

= 2âˆ«(t2 âˆ’ 1)/(t4 + 1)dt

= 2âˆ«(1 âˆ’ 1/t2)/[(t + 1/t)2 âˆ’ 2]dt

Let t + 1/t = y, so we have, (1 âˆ’ 1/t2)dt = dy

So, the equation becomes,

= 2âˆ«dy/(y2 âˆ’ 2)

= (2/2âˆš2) log |(y âˆ’ âˆš2)/(y + âˆš2)| + c

= (1/âˆš2) log |(t2 + 1 âˆ’ âˆš2t)/(t2 + 1 + âˆš2t)| + c

= (1/âˆš2) log |[x + 2 âˆ’ âˆš(2x + 2)]/[x + 2 + âˆš(2x + 2)]| + c

Question 8. âˆ«1/[(x âˆ’ 1)âˆš(x2 + 1)]dx

Solution:

We have,

âˆ«1/[(x âˆ’ 1)âˆš(x2 + 1)]dx

Let x âˆ’ 1 = 1/t, so we have dx = (âˆ’1/t2)dt

So, the equation becomes,

= âˆ’âˆ«(1/t2)/[(1/t)âˆš[(1 + 1/t)2 + 1]]dt

= âˆ’âˆ«dt/âˆš(2t2 + 2t + 1)

= âˆ’(1/âˆš2)âˆ«dt/âˆš(t2 + t + 1/2)

= âˆ’(1/âˆš2)âˆ«dt/âˆš[(t + 1/2)2 + 1/4]

= âˆ’(1/âˆš2) log |(t + 1/2) + âˆš[(t + 1/2)2 + 1/4]| + c

= âˆ’(1/âˆš2) log |(1/(x âˆ’ 1) + 1/2) + âˆš[(1/(xâˆ’1) + 1/2)2 + 1/4]| + c

Question 9. âˆ«1/[(x + 1)âˆš(x2 + x + 1)]dx

Solution:

We have,

âˆ«1/[(x + 1)âˆš(x2 + x + 1)]dx

Let x + 1 = 1/t, so we have dx = (âˆ’1/t2)dt

So, the equation becomes,

= âˆ’âˆ«(1/t2)/[(1/t)âˆš(1/t2 + 1/t âˆ’ 1)]dt

= âˆ’âˆ«dt/âˆš(1 + t âˆ’ t2)

= âˆ’âˆ«dt/âˆš[5/4 âˆ’ (1/4 âˆ’ t + t2)]

= âˆ’âˆ«dt/âˆš[5/4 âˆ’ (t âˆ’ 1/2)2]

= âˆ’ sinâˆ’1[(t âˆ’ 1/2)/(âˆš5/2)] + c

= âˆ’ sinâˆ’1[(2t âˆ’ 1)/âˆš5] + c

= âˆ’ sinâˆ’1[(1 âˆ’ x)/[âˆš5(x + 1)]] + c

Question 10. âˆ«1/[(x2 âˆ’ 1)âˆš(x2 + 1)]dx

Solution:

We have,

âˆ«1/[(x2 âˆ’ 1)âˆš(x2 + 1)]dx

Let x = 1/t, so we get, dx = (âˆ’1/t2)dt

So, the equation becomes,

= âˆ’âˆ«(1/t2)/[(1/t2 âˆ’ 1)âˆš(1/t2 + 1)]dt

= âˆ’âˆ«t/[(1 âˆ’ t2)âˆš(1 + t2)]dt

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

=> tdt = ydy

So, the equation becomes,

= âˆ«ydy/(y2 âˆ’ 2)y

= âˆ«dy/(y2 âˆ’ 2)

= (1/2âˆš2) log |(y âˆ’ âˆš2)/(y + âˆš2)| + c

= (1/2âˆš2) log |(y âˆ’ âˆš2)/(y + âˆš2)| + c

= (1/2âˆš2) log |[âˆš(1 + t2) âˆ’ âˆš2]/[âˆš(1 + t2) + âˆš2]| + c

= âˆ’(1/2âˆš2) log |[âˆš2x + âˆš(x2 + 1)]/[âˆš2x âˆ’ âˆš(x2 + 1)]| + c

Question 11. âˆ«x/[(x2 + 4)âˆš(x2 + 1)]dx

Solution:

We have,

âˆ«x/[(x2 + 4)âˆš(x2 + 1)]dx

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

=> xdx = tdt

So, the equation becomes,

= âˆ«t/(t2 + 3)(t)dt

= âˆ«dt/(t2 + 3)

= (1/âˆš3) tanâˆ’1(t/âˆš3) + c

= (1/âˆš3) tanâˆ’1[âˆš(x2 + 1)/âˆš3] + c

Question 12. âˆ«1/[(1 + x2)âˆš(1 âˆ’ x2)]dx

Solution:

We have,

âˆ«1/[(1 + x2)âˆš(1 âˆ’ x2)]dx

Let x = 1/t, so we get, dx = (âˆ’1/t2)dt

So, the equation becomes,

= âˆ’âˆ«(1/t2)/[(1/t2 + 1)âˆš(1 âˆ’ 1/t2)]dt

= âˆ’âˆ«t/[(t2 + 1)âˆš(t2 âˆ’ 1)]dt

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

=> tdt = ydy

So, the equation becomes,

= âˆ’âˆ«y/[(y2 + 2)(y)]dy

= âˆ’âˆ«1/(y2 + 2)dy

= âˆ’(1/âˆš2) tanâˆ’1(y/âˆš2) + c

= âˆ’(1/âˆš2) tanâˆ’1(âˆš(t2 âˆ’ 1)/âˆš2) + c

= âˆ’(1/âˆš2) tanâˆ’1(âˆš(1 âˆ’ x2)/âˆš2x) + c

Question 13. âˆ«1/[(2x2 + 3)âˆš(x2 âˆ’ 4)]dx

Solution:

We have,

âˆ«1/[(2x2 + 3)âˆš(x2 âˆ’ 4)]dx

Let x = 1/t, so we have dx = (âˆ’1/t2)dt

So, the equation becomes,

= âˆ’âˆ«(1/t2)/[(2/t2 + 3)âˆš(1/t2 âˆ’ 4)]dt

= âˆ’âˆ«t/[(2 + 3t2)âˆš(1âˆ’4t2)]dt

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

So, the equation becomes,

= (1/4) âˆ«y/[(11 âˆ’ 3y2)y/4]dy

= (1/3) âˆ«1/(11/3 âˆ’ y2)dy

= (1/2âˆš33) log |[y âˆ’ âˆš(11/3)]/[y + âˆš(11/3)]| + c

= (1/2âˆš33) log |[âˆš(1 âˆ’ 4t2) âˆ’ âˆš(11/3)]/[âˆš(1 + 4t2) + âˆš(11/3)]| + c

= (1/2âˆš33) log |[âˆš(11x) + âˆš(3x2 âˆ’ 12)]/[âˆš(11x) âˆ’ âˆš(3x2 âˆ’ 12)]| + c

Question 14. âˆ«x/[(x2 + 4)âˆš(x2 + 9)]dx

Solution:

We have,

âˆ«x/[(x2 + 4)âˆš(x2 + 9)]dx

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

=> xdx = ydy

So, the equation becomes,

= âˆ«y/[(y2 âˆ’ 5)y]dy

= âˆ«1/(y2 âˆ’ 5)dy

= (1/2âˆš5) log |(y âˆ’ âˆš5)/(y + âˆš5)| + c

= (1/2âˆš5) log |(âˆš(x2 + 9) âˆ’ âˆš5)/(âˆš(x2 + 9) + âˆš5)| + c

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