Question

`int_-1^1x^2/(1+x^2)  dx=` ______.

Options

• `(4+pi)/2`

• `(4-pi)/2`

• `(pi-4)/2`

• `4-pi/2`

Answer 1

`int_-1^1x^2/(1+x^2)  dx=underline((4-pi)/2)`.

Explanation:

Since `x^2/(1+x^2)` is an even function,

`int_-1^1x^2/(1+x^2)  dx=2int_0^1x^2/(1+x^2)  dx`

`=2int_0^1(x^2+1-1)/(1+x^2)  dx=2int_0^1(1-1/(1+x^2))  dx`

`=2[x-tan^-1(x)]_0^1`

`=2(1-pi/4)`

`=(4-pi)/2`