Question

Evaluate the following definite integrals:

`int_0^1 (1 - x^2)/(1 + x^2)^2  "d"x`

Answer 1

Let I = `int_0^1 (1 - x^2)/(1 + x^2)^2  "d"x`

= `int_0^1[2/(1 + x^2)^2 - (1 + x^2)/(1 + x^2)^2]"d"x`

I = `int_0^1 [2/(1 + x^2)^2 - 1/((1 + x^2))]"d"x`   ........(1)

I₁ = `int_0^1 2/(1 + x^2)^2  "d"x`

Put x = tan θ

dx = sec²θ dθ

x	0	1
θ	0	`pi/4`

= `2 int_0^(pi/4) (sec^2theta  "d"theta)/(1 + tan^2theta)^2`

= `2 int_0^(pi/4) (sec^2theta)/(sec^2theta)^2  "d"theta`

= `2 int_0^(pi/4) cos^2theta  "d"theta`

= `2 int_0^(pi/4) ((1 + cos 2theta)/2)  "d"theta`

= `(theta + (sin 2theta)/2)_0^(pi/4)`

= `pi/4 + 1/2`

I₂ = `int_0^1 1/(1 + x^2)  "d"x`

= `[tan^-1 x]_0^1`

= `pi/4`

(1) ⇒ I = `pi/4 + 1/2 - pi/4`

I = `1/2`