Evaluate the following :
`int_(0)^(1)x^(3)tan^(-1)xdx`

Explanation

Let `I=int_(0)^(1)x^(3).tan^(-1)xdx=int_(0)^(1)(tan^(-1)x).x^(3)dx`
`=[(tan^(-1)x)intx^(3)dx]_(0)^(1)-int_(0)^(1)[(d)/(dx)(tan^(-1)x)intx^(3)dx]dx`
`=[(tan^(-1)x).(x^(4))/(4)]_(0)^(1)-int_(0)^(1)(1)/(1+x^(2)).(x^(4))/(4)dx`
`=(1)/(4)[x^(4).tan^(-1)x]_(0)^(1)+(1)/(4)int_(0)^(1)(-x^(4))/(1+x^(2))dx`
`=(1)/(4)[tan^(-1)1-0]+(1)/(4)int_(0)^(1)((1-x^(4))-1)/(1+x^(2))dx`
`=(1)/(4).(pi)/(4)+(1)/(4)int_(0)^(1)((1-x^(4))/(1+x^(2))-(1)/(1+x^(2)))dx`
`=(pi)/(16)+(1)/(4)int_(0)^(1)[((1-x^(2))(1+x^(2)))/(1+x^(2))-(1)/(1+x^(2))]dx`
`=(pi)/(16)+(1)/(4)int_(0)^(1)[(1-x^(2))-(1)/(1+x^(2))]dx`
`=(pi)/(16)+(1)/(4)[x-(x^(3))/(3)-tan^(-1)x]_(0)^(1)`
`=(pi)/(16)+(1)/(4)[(1-(1)/(3)-tan^(-1)1)-(0-0-0)]`
`=(pi)/(16)+(1)/(4)((2)/(3)-(pi)/(4))`
`=(pi)/(16)+(1)/(6)-(pi)/(16)=(1)/(6).`