Evaluate: `intxcot^(-1)x\ dx`

Explanation

Given integral is,
`intxcot^(-1)x\ dx`
let`I= intxcot^(-1)x\ dx`
`=cot^-1x intxdx-int[(d(cot^-1x))/(dx)intxdx]dx`
`=cot^-1x*x^2/2+int1/(1+x^2)*x^2/2dx`
`=x^2/2cot^-1+1/2intx^2/(1+x^2)dx`
`=x^2/2cot^-1+1/2int((x^2+1)-1)/(1+x^2)dx`
`=x^2/2cot^-1-1/2int1/(1+x^2)dx+1/2intdx`
`=x^2/2cot^-1x-tan^-1x/2+x/2+c`