I can't seem to get the answer to
2xy^3/(x^2y^2+1)
Or you may make the substitution $\displaystyle x^{2}y^{2} = t$,
so that $\displaystyle dt = 2x^{2}ydy$
Therefore $\displaystyle dy = dt/(2x^{2}y)$
and $\displaystyle y^{2} = t/x^{2}$
Therefore the integrand now becomes, $\displaystyle \int(1/x^{3})(t/(t+1))\;dt$
This can be evaluated using standard forms.
$\displaystyle \displaystyle \int \dfrac{y^3}{x^2y^2+1} =\displaystyle \int 2x \dfrac{y}{x^2}dy - \displaystyle \int 2x \dfrac{y}{x^2 (x^2 y^2+1)} dy$
$\displaystyle =\displaystyle \int 2 \dfrac{y}{x} \;dy -\displaystyle \int 2 \dfrac{y}{x(x^2 y^2+1)} dy$
for the second one, use substution.
$\displaystyle u=x^2y^2+1 \; \therefore du = x^2(2y)\;dy$ and finish..