We know that arctan x is differentiable
u = arctan x
du/dx = 1/1+x^2
x^2dx = dv so v = x^3/3
now int udv = uv - int vdu/dx dx
= x^3(arctan x) - int (x^3/3(1+x^2) dx)
let 1+ x^ 2 = t
x^2 = (t-1)
2xdx = - dt
so we get x^3/1(1+x^2) dx
= x^2xdx/(1+x^2)/3
now you can proceed as
= (t-1) (-dt/6)/t...