What is the indefinite integral of $x^2\sqrt{1+x^2}$

Question

I get this but I don't know if it is correct. I used a reduction formula for $\tan^{2n}(x)\sec^{3}(x)$. Any help would be appreciated. My Final Answer:

$$\frac{\sqrt{x^2+1} x}{8}+\frac{\sqrt{x^2+1} x^3}{4}-\frac{\log|\sqrt{1+x^2}+x|}{2}+C$$

Answer 1

Solution should be:

$$\frac{\sqrt{x^2+1} x}{8}+\frac{\sqrt{x^2+1} x^3}{4}-\frac{\sinh^{-1}(x)}{8}+C$$

Considering that $\sinh^{-1}(x) = \log(x+\sqrt{1+x^2})$ your solution looks almost correct.

Answer 2

Hint: Let $x=\sinh t$, and use the fact that $\cosh^2t-\sinh^2t=1,~\sinh't=\cosh t,~\sinh2t=$ $=2\sinh t\cosh t$, and substitute $u=e^t$. Alternately, let $x=\tan y$, and use the fact that $\tan'y=$ $=1+\tan^2y=\sec^2y$, along with the Weierstrass substitution.