`int_0^1xtan^(- 1) x dx=`

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 `I = int_(0)^(1) x tan^(-1)x dx`
`= [ tan^(-1) x int x dx]_(0)^(1) - int_(0)^(1)((d)/(dx) tan^(-1)x intxdx) 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/8-1/1[1-pi/4] = pi/8 - 1/2 + pi/8 = pi/4 - 1/2`