<s>[INST] 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} $.
 [/INST] $$
\begin{aligned}
\int \frac{x^{3}}{\sqrt{x^{2}+1}} d x &=\int x^{2} d\left(\sqrt{x^{2}+1}\right) \\
&=x^{2} \sqrt{x^{2}+1}-2 \int x \sqrt{x^{2}+1} d x \\
&=x^{2} \sqrt{x^{2}+1}-\frac{2}{3}\left(x^{2}+1\right)^{\frac{3}{2}}+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left[3 x^{2}-2\left(x^{2}+1\right)\right]+C \\
&=\frac{\sqrt{x^{2}+1}}{3}\left(x^{2}-2\right)+C
\end{aligned}
$$
</s>