Evaluate: `int_0^1(x t a n^(-1)x)/((1+x^2)^(3//2))dx`

Explanation

Let `I=int_(0)^(1)(x tan^(-1)x)/((1+x^(2))^(3//2))dx=int_(0)^(1)(x tan^(-1)x)/(sqrt(1+x^(2))).(1)/(1+x^(2))dx`
Put `tan^(-1)x=t" "therefore (1)/(1+x^(2))dx=dt and x= tant`
When `x=0, t = tan^(-1)0=0`
When `x=1, t= tan^(-1)1=(pi)/(4)`
`therefore I=int_(0)^(pi//4)(tant.t)/(sqrt(1+tan^(2)t))dt =int_(0)^(pi//4)(t tant)/(sec t)dt`
`=int_(0)^(pi//4)t.(sint)/(cost).cos t dt=int_(0)^(pi//4)t sin t dt`
`=[t int sin t dt]_(0)^(pi//4)-int_(0)^(pi//4)[(d)/(dt)(t) int sin t dt] dt`
`=[t (-cost)]_(0)^(pi//4)-int_(0)^(pi//4)1.(-cos t)dt`
`=-[t cos t]_(0)^(pi//4)+int_(0)^(pi//4)cos tdt`
`=-((pi)/(4)cos.(pi)/(4)-0)+[sin t]_(0)^(pi//4)`
`=-(pi)/(4)xx(1)/(sqrt2)+sin.(pi)/(4)-sin0`
`=-(pi)/(4sqrt2)+(1)/(sqrt2)=(4-pi)/(4sqrt2)`.