Integral of $x^2 \sqrt{1 + x^2}$ How may one evaluate $\int x^2 \sqrt{x^2 + 1}\ dx$? I tried parts by integrating $\sqrt{x^2 + 1}$ but that seems to lead me down a rabbit hole of endless computations.
 A: Seeing $\sqrt{1 \pm x^2}$ is usually a sign to look at trig substitutions or hyperbolic trig substitutions that will let you simplify the square root without making other parts of the integrand worse.  In this case, consider $x=\sinh u$.  Then $x^2+1=\cosh^2 u$, and $dx=\cosh u \;du$, giving
$$
\int x^2\sqrt{x^2+1}\;dx=\int\sinh^2u\sqrt{\cosh^2u}\cosh u\;du=\int \sinh^2u\cosh^2u\;du.
$$
This can be evaluated using the double-angle formulas (and of course $u=\sinh^{-1}x$ must be substituted back in at the end).
A: Substitute $x=\tan(u)$ then we have $$\int x^2 \sqrt{x^2 + 1}\ dx$$
$$=\int \tan^2(u)\sec^3(u)du=\int(\sec^2(u)-1)\sec^{3}(u)du$$
$$=\int \sec^5(u)du-\int \sec^3(u)du.$$
Now applying the reduction formula $$I_{n}=\int \sec^{n}(x)dx=\frac{1}{n-1}\tan(x)\sec^{n-2}(x)+\frac{n-2}{n-1}I_{n-2}$$ gives $$=\int \sec^5(u)du-\int \sec^3(u)du$$
$$=\frac{1}{4}\tan(u)\sec^3(u)+\frac{3}{4}\int\sec^3(u)du-\int\sec^3(u)du$$
$$=\frac{1}{4}\tan(u)\sec^3{u}-\frac{1}{4}\big[\frac{1}{2}\tan(x)\sec(x)+\frac{1}{2}\int\sec(u)du\big]$$
$$=\frac{1}{4}\tan(u)\sec^3(u)-\frac{1}{8}\tan(u)\sec(u)-\frac{1}{8}\ln|\tan(u)+\sec(u)|+C$$
Finally since $u=\arctan(x)$, we have $\tan(u)=x$ and $\sec(u)=\sqrt{x^2+1},$ thus
$$\int x^2 \sqrt{x^2 + 1}\ dx=\frac{1}{4}x(x^2+1)^{\frac{3}{2}}-\frac{1}{8}x\sqrt{x^2+1}-\frac{1}{8}\ln|x+\sqrt{x^2+1}|+C.$$
A: You can use the hyperbolic sinus and cosinus, fulfilling $1 + \sinh^2(y) = \cosh^2(y)$.
$\sinh$ is monotone, so you can substitute
$x = \sinh(y),~ dx = \cosh(y)dy$.
It follows that:
$\int x^2\sqrt{1+x^2} dx = \int \sinh^2(y) \cosh^2(y) dy = \frac{1}{32} (\sinh^2(y) - 4y)$.
The latter integral can be solved by smart integration by parts.
A: $$
\int x^2\sqrt{x^2+1}\mathrm dx=\int(x^2+1)^{3/2}\mathrm dx-\int\sqrt{x^2+1}\mathrm dx
$$
Now, let's integrate the general case:
$$
\underbrace{\int(x^2+1)^{n/2}\mathrm dx}_{x=\sinh t}=\int\cosh^{n+1}t\mathrm dt
$$
Now, subsitute the original definition $\cosh t=(e^t+e^{-t})/2$, so
$$
\begin{aligned}
\int\cosh^{n+1}t\mathrm dt
&=2^{-n-1}\int(e^t+e^{-t})^{n+1}\mathrm dt \\
&=2^{-n-1}\sum_{k=0}^{n+1}\binom{n+1}k\int e^{(2k-n-1)t}\mathrm dt \\
&=2^{-n-1}\sum_{k=0}^{n+1}\binom{n+1}k{e^{(2k-n-1)t}\over2k-n-1}+C
\end{aligned}
$$
It can be verified that $e^t=\sqrt{1+\sinh^2t}+\sinh t=\sqrt{x^2+1}+x$, so we have
$$
I(n)\triangleq\int(x^2+1)^{n/2}\mathrm dx=2^{-n-1}\sum_{k=0}^{n+1}\binom{n+1}k{(\sqrt{x^2+1}+x)^{2k-n-1}\over2k-n-1}+C\tag1
$$
As a result, we obtain
$$
\int x^2\sqrt{x^2+1}\mathrm dx=I(3)-I(1)
$$
Plugging the (1) in should give desired result.
