Question

`int_0^1 x tan^-1x  dx` = ______ 

Options

• `pi/4 + 1/2`

• `pi/4 - 1/2`

• `1/2 - pi/4`

• `-pi/4 - 1/2`

Answer 1

`int_0^1 x tan^-1x  dx` = `underline(pi/4 - 1/2)`

Explanation:

`int_0^1 x tan^-1x  dx`

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

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

= `(pi/4 . 1/2 - 0) - 1/2[x - tan^-1x]_0^1`

= `pi/8 - 1/2[(1 - 0) - (pi/4 - 0)]`

= `pi/8 - 1/2 + pi/8 = pi/4 - 1/2`