Indefinite integral of $ \int \frac{x^3}{\sqrt{x^2+1}} \,\text{d}x$. Can you please provide any sort of hint or suggestion on how to find the following indefinite integral?
$$\int\frac{x^3}{\sqrt{x^2+1}}\text{d}x$$
I tried substituting everything but it didn't work. I also tried trigonometric substitution but I couldn't find any valid trigonometric identity for $ \sqrt{\cos^2(x)+1} $ or $ \sqrt{\sin^2(x)+1} $.
 A: We have 
$$\int \frac{x^3}{\sqrt{x^2 + 1}}\, dx = \int x \frac{(x^2 + 1) - 1}{\sqrt{x^2 + 1}}\, dx = \int x\sqrt{x^2 + 1}\, dx - \int \frac{x}{\sqrt{x^2 + 1}}\, dx.$$
Now use the $u$-substitution $u = x^2 + 1$ to get 
$$\frac{1}{2}\int \sqrt{u}\, du - \frac{1}{2}\int \frac{du}{\sqrt{u}} =... $$
A: Hint: Make the substitution $x^2+1=v^2$
The answer will come out as $\left(\dfrac{v^3}{3}-v+C\right)$ where C is the integration constant.
A: $$
\int\frac{x^2}{\sqrt{x^2+1}}\Big(x\,dx\Big) = \frac 1 2 \int \frac{u}{\sqrt{u+1}} \, du = \frac 1 2 \int \frac{w^2-1}{w}  \Big(2w\, dw\Big) = \cdots
$$
($w=\sqrt{u+1}$, so $w^2-1=u$ and then differentiating both sides of the last equality we get $2w\,dw=du$.)
A: $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{x^{2}+1}} d x &=\int x^{2} d\left(\sqrt{x^{2}+1}\right) \\
&=x^{2} \sqrt{x^{2}+1}-2 \int x \sqrt{x^{2}+1} d x \\
&=x^{2} \sqrt{x^{2}+1}-\frac{2}{3}\left(x^{2}+1\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left[3 x^{2}-2\left(x^{2}+1\right)\right]+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}-2\right)+C
\end{aligned}
$$
