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

Answer

log 2

Explanation

`int_(0)^(1)tan^(-1)((1)/(1-x+x^(2)))dx=int_(0)^(1)tan^(-1)[(1-x+x)/(1-x(1-x))]dx`
`=int_(0)^(1)[tan^(-1)(1-x)-tan^(-1)x]dx`
`=int_(0)^(1)tan^(-1)[1-(1-x)]dx+int_(0)^(1)tan^(-1)x dx`
`=2int_(0)^(1)tan^(-1)x dx [:' int_(0)^(a)f(x)=int_(0)^(a)f(a-x)dx]` . . . (i)
Now , `int_(0)^(1)tan^(-1)((1)/(1-x+x^(2)))dx`
`=int _(0)^(1)[(pi)/(2)-cot^(-1)((1)/(1-x+x^(2)))]dx`
` =(pi)/(2)-int_(0)^(1)tan^(-1)(1-x+x^(2))dx`
`:. int_(0)^(1)tan^(-1)(1-x+x^(2))dx= (pi)/(2)- int_(0)^(1)tan^(-1)(1)/((1-x+x^(2)))dx`
`= (pi)/(2)-2I_(2)`
Where , `I_(1)=int_(0)^(1)tan^(-1)xdx=[xtan^(-1)x]_(0)^(1)-int_(0)^(1)(xdx)/(1+x^(2))`
`=(pi)/(4)-(1)/(2) [log (1+x^(2))] _(0)^(1) =(pi)/(4)-(1)/(2)log 2`
`:. int _(0)^(1)tan^(-1)(1-x+x^(2))dx = (pi)/(2)-2 ((pi)/(4)-(1)/(2)log 2 )=log 2`