<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] 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 
$$\frac{1}{2}\int \sqrt{u}\, du - \frac{1}{2}\int \frac{du}{\sqrt{u}} =... $$
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