Evaluate the integral using u substitution
((y^3)/((1-y^2)^2))dy
u = 1 - y^2
du = -2 y dy or dy = du/-2y
((y^3)/(u^2))[du/-2y]
(-1/2)y^2 du/u^2
but y^2 = 1-u
so
-(1/2)(1-u) du/u^2
= (u/2)du/u^2 - (1/2)du/u^2
= (1/2) du/u - .5 u^-2 du
finish and check my arithmetic :)