Evaluate the integrals `int0 1sin^(-1)((2x)/(1+x^2))dx`

Explanation

`"Let " I = int_(0)^(1) sin^(-1) ((2x)/(1+x^(2)))dx`
`" Let " tan 0 =x rArr sec^(2) 0 d0 =dx`
` x=0 rArr 0=0`
`" and " x=1 rArr 0=(pi)/(4)`
`:. I=int_(0)^(pi//4) sin^(-1) ((2 tan 0)/(1+tan^(2)0)) sec^(2) 0 d0`
` =2int_(0)^(pi//4) underset(I)(0)sec^(2) underset(II)(0)d0`
` I=2 [0int sec^(2)0d0 -int {((d)/(d0)0)} int sec^(2) 0 d0 ]_(0)^(pi//4)`
`=2[0 tan 0 -int tan 0 d0 ]_(0)^(pi//4)`
`=2{((pi)/(4) tan .(pi)/(4)+log |cos.(pi)/(4)|)`
`(0+log |cos 0|)}( :. log 1 =0)`
=(pi)/(2) +2log.(1)/(sqrt(2))-0`
=(pi)/(2) +2 log 2^(-1//2)`
`=(pi)/(2)-2 xx(1)/(2) log 2`
`=(pi)/(2) -log 2`