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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.31

• Last Updated : 07 Apr, 2021

### Question 1. âˆ«(x2 + 1)/(x4 + x2 + 1)dx

Solution:

We have,

âˆ«(x2 + 1)/(x4 + x2 + 1)dx

= âˆ«x2(1 + 1/x2)/x2(x2 + 1 + 1/x2)dx

= âˆ«(1 + 1/x2)/(x2 + 1 + 1/x2)dx

= âˆ«(1 + 1/x2)/(x2 + 1/x2 âˆ’ 2 + 2 + 1)dx

= âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 3]dx

Let x âˆ’ 1/x = t.

Differentiating both sides, we get,

(1 + 1/x2)dx = dt

So, our equation becomes,

= âˆ«1/(t2 + 3)dt

= (1/âˆš3) tan-1(t/âˆš3) + c

= (1/âˆš3) tan-1[(x2 âˆ’ 1)/âˆš3x] + c

### Question 2. âˆ«âˆš(cotÎ¸)dÎ¸

Solution:

We have,

âˆ«âˆš(cotÎ¸)dÎ¸

Let cotÎ¸ = x2

Differentiating both sides, we get,

=> âˆ’cosec2Î¸dÎ¸ = 2xdx

=> dÎ¸ = âˆ’2xdx/cosec2Î¸

=> dÎ¸ = âˆ’2xdx/(1 + cot2Î¸)

=> dÎ¸ = âˆ’2xdx/(1 + x4)

So, our equation becomes,

= âˆ’âˆ«2x2/(1 + x4)]dx

= âˆ’âˆ«2/(x2 + 1/x2)dx

= âˆ’âˆ«(1 + 1/x2 + 1 âˆ’ 1/x2)/(x2 + 1/x2 + 2 âˆ’ 2)dx

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

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x + 1/x = y, so we get, (1 âˆ’ 1/x2)dx = dy

So, our equation becomes,

= âˆ’âˆ«dt/(t2 + 2) âˆ’ âˆ«dy/(y2 âˆ’ 2)

= âˆ’(1/âˆš2) tan-1[t/âˆš2] âˆ’ (1/2âˆš2) log |(y âˆ’ âˆš2)/(y + âˆš2)| + c

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

= âˆ’(1/âˆš2) tan-1[(cotÎ¸ âˆ’ 1)/âˆš(2cotÎ¸)] âˆ’ (1/2âˆš2) log |(cotÎ¸ + 1 âˆ’ âˆš(2cotÎ¸))/(cotÎ¸ + 1 + âˆš(2cotÎ¸))| + c

### Question 3. âˆ«(x2 + 9)/(x4 + 81)dx

Solution:

We have,

âˆ«(x2 + 9)/(x4 + 81)dx

= âˆ«x2(1 + 9/x2)/x2(x2 + 81/x2)dx

= âˆ«(1 + 9/x2)/[(x âˆ’ 9/x)2 + 18]dx

Let xâˆ’9/x = t, so we have, (1 + 9/x2)dx = dt

So, our equation becomes,

= âˆ«dt/(t2 + 18)

= (1/3âˆš2) tan-1(t/3âˆš2) +c

= (1/3âˆš2) tan-1[(x2 âˆ’ 9)/3âˆš2x] + c

### Question 4. âˆ«1/(x4 + x2 + 1)dx

Solution:

We have,

âˆ«1/(x4 + x2 + 1)dx

= âˆ«(1/x2)/(x2 + 1 + 1/x2)dx

= (1/2)âˆ«(1 + 1/x2 âˆ’ 1 + 1/x2)/(x2 + 1 + 1/x2)dx

= (1/2)âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 3]dx âˆ’ (1/2)âˆ«(1 âˆ’ 1/x2)/(x + 1/x)2âˆ’ 1)dx

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt and

Let x + 1/x = y, so we get, (1 âˆ’ 1/x2)dx = dy.

So, our equation becomes,

= (1/2)âˆ«dt/(t2 + 3) âˆ’ (1/2)âˆ«dy/(y2 âˆ’ 1)

= (1/2âˆš3) tan-1(t/âˆš3) âˆ’ (1/4) log |(y âˆ’ 1)/(y + 1)| + c

= (1/2âˆš3) tan-1[(x2 âˆ’ 1)/âˆš3x] âˆ’ (1/4) log |(x2 + 1 âˆ’ x)/(x2 + 1 + x)| + c

### Question 5. âˆ«(x2 âˆ’ 3x + 1)/(x4 + x2 + 1)dx

Solution:

We have,

âˆ«(x2 âˆ’ 3x + 1)/(x4 + x2 + 1)dx

= âˆ«(1 âˆ’ 3/x + 1/x2)/(x2 + 1 + 1/x2)dx

= âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 3]dx âˆ’ âˆ«3x/(x4 + x2 + 1)dx

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x2 = y, so we get, 2xdx = dy

So, our equation becomes,

= âˆ«dt/(t2 + 3) âˆ’ (3/2)âˆ«dy/(y2 + y + 1)

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

= (1/âˆš3) tan-1(t/âˆš3) âˆ’ (3/2)(2/âˆš3)tan-1[(y + 1/2)/(âˆš3/2)] + c

= (1/âˆš3) tan-1(t/âˆš3) âˆ’ âˆš3tan-1[(2y + 1)/âˆš3] + c

= (1/âˆš3) tan-1[(x2 âˆ’ 1)/âˆš3] âˆ’ âˆš3tan-1[(2x2 + 1)/âˆš3] + c

### Question 6. âˆ«(x2 + 1)/(x4 âˆ’ x2 + 1)dx

Solution:

We have,

âˆ«(x2 + 1)/(x4 âˆ’ x2 + 1)dx

= âˆ«(1 + 1/x2)/(x2 âˆ’ 1 + 1/x2)dx

= âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 1]dx

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt

So, the equation becomes,

= âˆ«dt/(t2 + 1)

= tanâˆ’1t + c

= tanâˆ’1[(x2 âˆ’ 1)/x] + c

### Question 7. âˆ«(x2 âˆ’ 1)/(x4 + 1)dx

Solution:

We have,

âˆ«(x2 âˆ’ 1)/(x4 + 1)dx

= âˆ«(1 âˆ’ 1/x2)/(x2 + 1/x2)dx

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

Let x + 1/x = t, so we get, (1 âˆ’ 1/x2)dx = dt

So, the equation becomes,

= âˆ«dt/(t2 âˆ’ 2)

= (1/2âˆš2) log |(t âˆ’ âˆš2)/(t + âˆš2)| + c

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

### Question 8. âˆ«(x2 + 1)/(x4 + 7x2 + 1)dx

Solution:

We have,

âˆ«(x2 + 1)/(x4 + 7x2 + 1)dx

= âˆ«(1 + 1/x2)/(x2 + 7 + 1/x2)dx

= âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 9]dx

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt

So, the equation becomes,

= âˆ«dt/(t2 + 9)

= (1/3)tanâˆ’1(t/3) + c

= (1/3)tanâˆ’1[(x2 âˆ’ 1)/3x] + c

### Question 9. âˆ«(x âˆ’ 1)2/(x4 + x2 + 1)dx

Solution:

We have,

âˆ«(x âˆ’ 1)2/(x4 + x2 + 1)dx

= âˆ«(x2 âˆ’ 2x + 1)/(x4 + x2 + 1)dx

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

= âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 3]dx âˆ’ âˆ«2x/(x4 + x2 + 1)dx

Let x âˆ’ 1/x = t, so we get, (1 + 1/x2)dx = dt

Let x2 = y, we get, 2xdx = dy

So, our equation becomes,

= âˆ«dt/(t2 + 3) âˆ’ âˆ«dy/(y2 + y + 1)

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

= (1/âˆš3) tanâˆ’1[(x2 âˆ’ 1)/âˆš3x] âˆ’ (2/âˆš3) tanâˆ’1[(2x2 + 1)/âˆš3] + c

### Question 10. âˆ«1/(x4 + 3x2 + 1)dx

Solution:

We have,

âˆ«1/(x4 + 3x2 + 1)dx

= âˆ«(1/x2)/(x2 + 3 + 1/x2)dx

= (1/2)âˆ«(1 + 1/x2 âˆ’ 1 + 1/x2)/(x2 + 3 + 1/x2)dx

= (1/2)âˆ«(1 + 1/x2)/[(x âˆ’ 1/x)2 + 5]dx âˆ’ (1/2)âˆ«(1 âˆ’ 1/x2)/[(x + 1/x)2 + 1]dx

Let x âˆ’ 1/x = t, we get, (1 + 1/x2)dx = dt

Let x + 1/x = y, we get, (1 âˆ’ 1/x2)dx = dy

So, the equation becomes,

= (1/2)âˆ«dt/(t2 + 5) âˆ’ (1/2)âˆ«dy/(y2 + 1)

= (1/2âˆš5) tanâˆ’1(t/âˆš5) âˆ’ (1/2) tanâˆ’1y + c

= (1/2âˆš5) tanâˆ’1[(x2 âˆ’ 1)/âˆš5x] âˆ’ (1/2) tanâˆ’1[(x2 + 1)/x] + c

### Question 11. âˆ«1/(sin4x + sin2x cos2x + cos4x)dx

Solution:

We have,

âˆ«1/(sin4x + sin2x cos2x + cos4x)dx

= âˆ«(1/cos4x)/[(sin4x + sin2x cos2x + cos4x)/(cos4x)]dx

= âˆ«(sec4x)/(tan4x + tan2x + 1)dx

= âˆ«[(1 + tan2x)sec2x]/(tan4x + tan2x + 1)dx

Let tan x = t, so we get sec2xdx = dt

So, the equation becomes,

= âˆ«(1 + t2)/(t4 + t2 + 1)dt

= âˆ«(1 + 1/t2)/(t2 + 1/t2 + 1)dt

= âˆ«(1 + 1/t2)/[(t âˆ’ 1/t)2 + 3]dt

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

So, the equation becomes,

= âˆ«dy/(y2 + 3)

= (1/âˆš3) tanâˆ’1(y/âˆš3) + c

= (1/âˆš3) tanâˆ’1[(tâˆ’1/t)/âˆš3] + c

= (1/âˆš3) tanâˆ’1[(tan x âˆ’ 1/tan x)/âˆš3] + c

= (1/âˆš3) tanâˆ’1[(tan x âˆ’ cot x)/âˆš3] + c

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