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

### Evaluate the following integrals.

### Question 1. âˆ«(x^{2 }+ 1)/(x^{4 }+ x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«(x

^{2 }+ 1)/(x^{4 }+ x^{2 }+ 1)dx= âˆ«x

^{2}(1 + 1/x^{2})/x^{2}(x^{2 }+ 1 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/(x^{2 }+ 1 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/(x^{2 }+ 1/x^{2 }âˆ’ 2 + 2 + 1)dx= âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 3]dxLet x âˆ’ 1/x = t.

Differentiating both sides, we get,

(1 + 1/x

^{2})dx = dtSo, our equation becomes,

= âˆ«1/(t

^{2 }+ 3)dt= (1/âˆš3) tan

^{-1}(t/âˆš3) + c= (1/âˆš3) tan

^{-1}[(x^{2 }âˆ’ 1)/âˆš3x] + c

### Question 2. âˆ«âˆš(cotÎ¸)dÎ¸

**Solution:**

We have,

âˆ«âˆš(cotÎ¸)dÎ¸

Let cotÎ¸ = x

^{2}Differentiating both sides, we get,

=> âˆ’cosec

^{2}Î¸dÎ¸ = 2xdx=> dÎ¸ = âˆ’2xdx/cosec

^{2}Î¸=> dÎ¸ = âˆ’2xdx/(1 + cot

^{2}Î¸)=> dÎ¸ = âˆ’2xdx/(1 + x

^{4})So, our equation becomes,

= âˆ’âˆ«2x

^{2}/(1 + x^{4})]dx= âˆ’âˆ«2/(x

^{2 }+ 1/x^{2})dx= âˆ’âˆ«(1 + 1/x

^{2 }+ 1 âˆ’ 1/x^{2})/(x^{2 }+ 1/x^{2 }+ 2 âˆ’ 2)dx= âˆ’âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 2]dx âˆ’ âˆ«(1 âˆ’ 1/x^{2})/[(x + 1/x)^{2 }âˆ’ 2]dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dtLet x + 1/x = y, so we get, (1 âˆ’ 1/x

^{2})dx = dySo, our equation becomes,

= âˆ’âˆ«dt/(t

^{2 }+ 2) âˆ’ âˆ«dy/(y^{2 }âˆ’ 2)= âˆ’(1/âˆš2) tan

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

^{-1}[(x^{2 }âˆ’ 1)/âˆš2x] âˆ’ (1/2âˆš2) log |(x^{2 }+ 1 âˆ’ âˆš2x)/(x^{2 }+ 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«(x^{2 }+ 9)/(x^{4 }+ 81)dx

**Solution:**

We have,

âˆ«(x

^{2 }+ 9)/(x^{4 }+ 81)dx= âˆ«x

^{2}(1 + 9/x^{2})/x^{2}(x^{2 }+ 81/x^{2})dx= âˆ«(1 + 9/x

^{2})/[(x âˆ’ 9/x)^{2 }+ 18]dxLet xâˆ’9/x = t, so we have, (1 + 9/x

^{2})dx = dtSo, our equation becomes,

= âˆ«dt/(t

^{2 }+ 18)= (1/3âˆš2) tan

^{-1}(t/3âˆš2) +c= (1/3âˆš2) tan

^{-1}[(x^{2 }âˆ’ 9)/3âˆš2x] + c

### Question 4. âˆ«1/(x^{4 }+ x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«1/(x

^{4 }+ x^{2 }+ 1)dx= âˆ«(1/x

^{2})/(x^{2 }+ 1 + 1/x^{2})dx= (1/2)âˆ«(1 + 1/x

^{2 }âˆ’ 1 + 1/x^{2})/(x^{2 }+ 1 + 1/x^{2})dx= (1/2)âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 3]dx âˆ’ (1/2)âˆ«(1 âˆ’ 1/x^{2})/(x + 1/x)^{2}âˆ’ 1)dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dt andLet x + 1/x = y, so we get, (1 âˆ’ 1/x

^{2})dx = dy.So, our equation becomes,

= (1/2)âˆ«dt/(t

^{2 }+ 3) âˆ’ (1/2)âˆ«dy/(y^{2 }âˆ’ 1)= (1/2âˆš3) tan

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

^{-1}[(x^{2 }âˆ’ 1)/âˆš3x] âˆ’ (1/4) log |(x^{2 }+ 1 âˆ’ x)/(x^{2 }+ 1 + x)| + c

### Question 5. âˆ«(x^{2 }âˆ’ 3x + 1)/(x^{4 }+ x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«(x

^{2 }âˆ’ 3x + 1)/(x^{4 }+ x^{2 }+ 1)dx= âˆ«(1 âˆ’ 3/x + 1/x

^{2})/(x^{2 }+ 1 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 3]dx âˆ’ âˆ«3x/(x^{4 }+ x^{2 }+ 1)dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dtLet x

^{2}= y, so we get, 2xdx = dySo, our equation becomes,

= âˆ«dt/(t

^{2 }+ 3) âˆ’ (3/2)âˆ«dy/(y^{2 }+ y + 1)= âˆ«dt/(t

^{2 }+ 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}[(x^{2 }âˆ’ 1)/âˆš3] âˆ’ âˆš3tan^{-1}[(2x^{2 }+ 1)/âˆš3] + c

### Question 6. âˆ«(x^{2 }+ 1)/(x^{4 }âˆ’ x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«(x

^{2 }+ 1)/(x^{4 }âˆ’ x^{2 }+ 1)dx= âˆ«(1 + 1/x

^{2})/(x^{2 }âˆ’ 1 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 1]dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dtSo, the equation becomes,

= âˆ«dt/(t

^{2 }+ 1)= tan

^{âˆ’1}t + c= tan

^{âˆ’1}[(x^{2 }âˆ’ 1)/x] + c

### Question 7. âˆ«(x^{2 }âˆ’ 1)/(x^{4 }+ 1)dx

**Solution:**

We have,

âˆ«(x

^{2 }âˆ’ 1)/(x^{4 }+ 1)dx= âˆ«(1 âˆ’ 1/x

^{2})/(x^{2 }+ 1/x^{2})dx= âˆ«(1 âˆ’ 1/x

^{2})/[(x + 1/x)^{2 }âˆ’ 2]dxLet x + 1/x = t, so we get, (1 âˆ’ 1/x

^{2})dx = dtSo, the equation becomes,

= âˆ«dt/(t

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

= (1/2âˆš2) log |(x

^{2 }+ 1 âˆ’ âˆš2x)/(x^{2 }+ 1 + âˆš2x)| + c

### Question 8. âˆ«(x^{2 }+ 1)/(x^{4 }+ 7x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«(x

^{2 }+ 1)/(x^{4 }+ 7x^{2 }+ 1)dx= âˆ«(1 + 1/x

^{2})/(x^{2 }+ 7 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 9]dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dtSo, the equation becomes,

= âˆ«dt/(t

^{2 }+ 9)= (1/3)tan

^{âˆ’1}(t/3) + c= (1/3)tan

^{âˆ’1}[(x^{2 }âˆ’ 1)/3x] + c

### Question 9. âˆ«(x âˆ’ 1)^{2}/(x^{4 }+ x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«(x âˆ’ 1)

^{2}/(x^{4 }+ x^{2 }+ 1)dx= âˆ«(x

^{2 }âˆ’ 2x + 1)/(x^{4 }+ x^{2 }+ 1)dx= âˆ«(1 âˆ’ 2/x + 1/x

^{2})/(x^{2 }+ 1 + 1/x^{2})dx= âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 3]dx âˆ’ âˆ«2x/(x^{4 }+ x^{2 }+ 1)dxLet x âˆ’ 1/x = t, so we get, (1 + 1/x

^{2})dx = dtLet x

^{2}= y, we get, 2xdx = dySo, our equation becomes,

= âˆ«dt/(t

^{2 }+ 3) âˆ’ âˆ«dy/(y^{2 }+ y + 1)= âˆ«dt/(t

^{2 }+ 3) âˆ’ âˆ«dy/[(y + 1/2)^{2 }+ 3/4]= (1/âˆš3) tan

^{âˆ’1}[(x^{2 }âˆ’ 1)/âˆš3x] âˆ’ (2/âˆš3) tan^{âˆ’1}[(2x^{2 }+ 1)/âˆš3] + c

### Question 10. âˆ«1/(x^{4 }+ 3x^{2 }+ 1)dx

**Solution:**

We have,

âˆ«1/(x

^{4 }+ 3x^{2 }+ 1)dx= âˆ«(1/x

^{2})/(x^{2 }+ 3 + 1/x^{2})dx= (1/2)âˆ«(1 + 1/x

^{2 }âˆ’ 1 + 1/x^{2})/(x^{2 }+ 3 + 1/x^{2})dx= (1/2)âˆ«(1 + 1/x

^{2})/[(x âˆ’ 1/x)^{2 }+ 5]dx âˆ’ (1/2)âˆ«(1 âˆ’ 1/x^{2})/[(x + 1/x)^{2 }+ 1]dxLet x âˆ’ 1/x = t, we get, (1 + 1/x

^{2})dx = dtLet x + 1/x = y, we get, (1 âˆ’ 1/x

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

= (1/2)âˆ«dt/(t

^{2 }+ 5) âˆ’ (1/2)âˆ«dy/(y^{2 }+ 1)= (1/2âˆš5) tan

^{âˆ’1}(t/âˆš5) âˆ’ (1/2) tan^{âˆ’1}y + c= (1/2âˆš5) tan

^{âˆ’1}[(x^{2 }âˆ’ 1)/âˆš5x] âˆ’ (1/2) tan^{âˆ’1}[(x^{2 }+ 1)/x] + c

### Question 11. âˆ«1/(sin^{4}x + sin^{2}x cos^{2}x + cos^{4}x)dx

**Solution:**

We have,

âˆ«1/(sin

^{4}x + sin^{2}x cos^{2}x + cos^{4}x)dx= âˆ«(1/cos

^{4}x)/[(sin^{4}x + sin^{2}x cos^{2}x + cos^{4}x)/(cos^{4}x)]dx= âˆ«(sec

^{4}x)/(tan^{4}x + tan^{2}x + 1)dx= âˆ«[(1 + tan

^{2}x)sec^{2}x]/(tan^{4}x + tan^{2}x + 1)dxLet tan x = t, so we get sec

^{2}xdx = dtSo, the equation becomes,

= âˆ«(1 + t

^{2})/(t^{4 }+ t^{2 }+ 1)dt= âˆ«(1 + 1/t

^{2})/(t^{2 }+ 1/t^{2 }+ 1)dt= âˆ«(1 + 1/t

^{2})/[(t âˆ’ 1/t)^{2 }+ 3]dtLet t âˆ’ 1/t = y, so we get, (1 + 1/t

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

= âˆ«dy/(y

^{2 }+ 3)= (1/âˆš3) tan

^{âˆ’1}(y/âˆš3) + c= (1/âˆš3) tan

^{âˆ’1}[(tâˆ’1/t)/âˆš3] + c= (1/âˆš3) tan

^{âˆ’1}[(tan x âˆ’ 1/tan x)/âˆš3] + c= (1/âˆš3) tan

^{âˆ’1}[(tan x âˆ’ cot x)/âˆš3] + c

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