`int_(0)^(1)xtan^(-1)xdx=`

Options

A. `(pi)/(4)+(1)/(2)`
B. `(pi)/(4)-(1)/(2)`
C. `(1)/(2)-(pi)/(4)`
D. `-(pi)/(4)-(1)/(2)`

Answer

B

Explanation

let `l=int_(0)^(1)xtan^(-1)xdx`
`=[tan^(-1)xintxdx]_(0)^(1)-int_(0)^(1)((d)/(dx)tan^(-1)x int xdx)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)/(2)int_(0)^(1)(1+x^(2)-1)/(1+x^(2))dx`
`=(1)/(2)(pi)/(4)-(1)/(2)int_(0)^(1)(1-(1)/(1+x^(2)))dx`
`=(pi)/(8)-(1)/(2)[x-tan^(-1)x]_(0)^(1)`
`=(pi)/(8)-(1)/(2)[1-tan^(-1)1-0+0]`
`=(pi)/(8)-(1)/(2)[1-(pi)/(4)]`
`=(pi)/(8)-(1)/(2)+(pi)/(8)=(pi)/(4)-(1)/(2)`