Question

`int_0^1 x tan^-1 x  dx` = ______.

Options

• `π/4 + 1/2`

• `π/4 - 1/2`

• `1/2 - π/4`

• `-π/4 - 1/2`

Answer 1

`int_0^1 x tan^-1 x  dx` = `underlinebb(π/4 - 1/2)`.

Explanation:

`int_0^1 x tan^-1 x  dx`

= `[tan^-1x intx  dx]_0^1 - int_0^1(d/dx(tan^-1x)intx  dx)dx`

= `[tan^-1 x x^2/2]_0^1 - int_0^1(1/(1 + x^2). x^2/2)dx`

= `(1/2 tan^-1 1 - 0) - 1/2int_0^1(1 + x^2 - 1)/(1 + x^2)dx`

= `1/2(π/4) - 1/2int_0^1(1 - 1/(1 + x^2))dx`

= `π/8 - 1/2[x - tan^-1x]_0^1`

= `π/8 - 1/2[1 - tan^-1 1 - 0 + 0]`

= `π/8 - 1/2[1 - π/4]`

= `π/4 - 1/2`