Question

Evaluate the integral using u substitution

((y^3)/((1-y^2)^2))dy

Answer 1

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 :)