Prove that `int_0^1tan^(-1)(1/(1-x+x^2))dx=2int_0^1tan^(-1)x dx`. Hence or otherwise, evaluate the integral `int_0^1tan^(-1)(1-x+x^2)dx`

Explanation

`int_(0)^(1)"tan"^(-1)1/(1-x+x^(2))dx=int_(0)^(1)"tan"^(-1)(x+(1-x))/(1-x(1-x))dx`
`=int_(0)^(1)[tan^(-1)x+tan^(-1)(1-x)]dx`
`=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)(1-x)dx`
`=int_(0)^(1)tan^(-1)xdx+int_(0)^(1)tan^(-1)tan^(-1)[1-(1-x)]dx`
`=2int_(0)^(1)tan^(-1)x dx`................1
Now, `I=int_(0)^(1)tan^(-1)(1-x+x^(2))dx`
`=int_(0)^(1)cot^(-1)(1/(1-x+x^(2)))dx`
`=int_(0)^(1)[(pi)/2-tan^(-1)(1/(1-x+x^(2)))]dx`
`=(pi)/2-2int_(0)^(1)x dx`[from equation 1]
`=(pi)/2-2{x tan^(-1) x-1/2 log(1+x^(2))}_(0)^(1)`
`=log_(e)2`