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

1.	Evaluate `int_0^1(tan^(-1)x)/(1+x^2)dx`
Answer» let `tan^-1 x = t` then `dx/(1+x^2) = dt` when `x=0, t=tan^-1 0 = 0` when `x=1, t=tan^-1 1 = pi/4` putting it in the given equation I `= int_0^(pi/4) t dt ` `= [t^2/2]= 1/2 [t^2]` `= 1/2 *(pi/4)^2 - 0)` `= pi^2/32` answer

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