Question

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

विकल्प

• `1 - pi/4`

• `1 + pi/4`

• `1 + (3pi)/4`

• `1 - (3pi)/4`

Answer 1

`1 - (3pi)/4`

Explanation:

We have,

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

I = `int_0^1 ((x^2 + 1)/(x^2 + 1)) "dx" - 3int "dx"/(x^2 + 1)`

I = `[x]_0^1 - 3 [tan^-1 x]_0^1`

I = 1 - 3(tan^-1 1 - tan^-1 0)

I = 1 - `(3 pi)/4`