Inverse Trig & Trig Sub Can someone explain to me how to solve this using inverse trig and trig sub?
$$\int\frac{x^3}{\sqrt{1+x^2}}\, dx$$
Thank you. 

You can also use integration by part: let $u=x^2$ and $dv=\frac{x}{\sqrt{1+x^2}}$ then you will have:
\begin{align*}
\int udv&=uv-\int vdu\\
&=x^2\sqrt{1+x^2}-\int2x\sqrt{1+x^2}\,dx\\
&=x^2\sqrt{1+x^2}-\frac{2}{3}(1+x^2)^{\frac{3}{2}}+C
\end{align*}
where the last integral was solved by substitution $\ u=1+x^2$
