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. 
 A: Set $u=1+x^2$. Then $\dfrac{du}{dx}=2x$ and so,
$\displaystyle\int \dfrac{x^3}{\sqrt{1+x^2}}dx=\int \dfrac12\dfrac{u-1}{\sqrt{u}}du=\dfrac12\int (u^{1/2}-u^{-1/2})\, du$.
A: $$ \frac{x^3}{\sqrt{1+x^2}}=\frac{x^3+x-x}{\sqrt{1+x^2}}=x\sqrt{1+x^2}-\frac x{\sqrt{1+x^2}}$$
Set $1+x^2=u$ in each case
A: 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 $\ u=1+x^2$
A: You can use hyperbolic substitution, i.e. let $x=\sinh t$ then
\begin{align*}
\int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \sinh^3 t\, dt\\
&=\int (\cosh^2t -1)\sinh t\, dt\\
&=\frac{\cosh^3 t}{3}-\cosh t+C\\
&=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C
\end{align*}
Also you can use triangle substitution : $x=\tan \theta$
\begin{align*}
\int\frac{x^3}{\sqrt{1+x^2}}\, dx&=\int \tan^3\theta\sec\theta\, d\theta\\
&=\int (\sec^2\theta-1)\sec\theta\tan\theta\, d\theta\\
&=\frac{\sec^3\theta}{3}-\sec\theta+C\\
&=\frac{(\sqrt{1+x^2})^3}{3}-\sqrt{1+x^2}+C
\end{align*}
