How to calculate $\int_{0}^{1} x^2 \sqrt{1+x^2} dx$? I'm trying to calculate the following integral:
$\int_{0}^{1} x^2 \sqrt{1+x^2} dx$
I tried solving by parts but i'm getting nowhere close. I feel like some substitution will be good here, however neither $x=\cos(u)$ nor $x=\sin(u)$ get me anywhere.
 A: Hint:
Try substituting $x = \sinh u$:
$$\int_0^1 x^2\sqrt{1+x^2}\,dx = \begin{vmatrix} x = \sinh u \\ dx = \cosh u\,du\end{vmatrix} = \int_0^{\operatorname{Arsinh 1}} \sinh^2u\cosh^2u\,du 
$$
This simplifies to just some exponential functions which should be easy to solve:
$$\int_0^{\operatorname{Arsinh 1}} \sinh^2u\cosh^2u\,du = \frac1{16} \int_0^{\operatorname{Arsinh 1}} (e^{2u}+e^{-2u} - 2) (e^{2u}+e^{-2u} + 2)\,du$$
A: First let $x = \tan u$, we get
\begin{equation}
 {\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
 ={\displaystyle\int}\sec^2\left(u\right)\tan^2\left(u\right)\sqrt{\tan^2\left(u\right)+1}\,\mathrm{d}u
\end{equation}
Using the fact that
    $\tan^2 u + 1 =\sec^2 u$
we get
\begin{equation}
 {\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
 ={\displaystyle\int}\sec^3\left(u\right)\tan^2\left(u\right)\,\mathrm{d}u
 ={\displaystyle\int}\sec^3\left(u\right)\left(\sec^2\left(u\right)-1\right)\,\mathrm{d}u
 =
 A_5 - A_3
\end{equation}
Expand to get
\begin{equation}
 {\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
 ={\displaystyle\int}\sec^5\left(u\right)\,\mathrm{d}u-{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u
\end{equation}
Lets work with $A_5$, using the reduction formula, 
\begin{equation}
\small{{\displaystyle\int}\sec^{n}\left(u\right)\,\mathrm{d}u={{\dfrac{n-2}{n-1}}}\int \sec^{n-2}\left(u\right)\,\mathrm{d}u+\dfrac{\sec^{n-2}\left(u\right)\tan\left(u\right)}{n-1}}
\end{equation}
which gives us 
\begin{equation}
 A_5 =\dfrac{\sec^3\left(u\right)\tan\left(u\right)}{4}+{{\dfrac{3}{4}}}A_3
\end{equation}
and 
\begin{equation}
 A_3 =\dfrac{\sec\left(u\right)\tan\left(u\right)}{2}+\frac{1}{2} A_1
\end{equation}
Now $A_1$ is easy and is known to be $A_1 = \ln\left(\tan\left(u\right)+\sec\left(u\right)\right)$. So plugging back upwards we get
\begin{equation}
 A_3 =\dfrac{\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{2}+\dfrac{\sec\left(u\right)\tan\left(u\right)}{2}
\end{equation}
and. hence the original integral becomes
\begin{equation}
 {\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
 =-\dfrac{\ln\left(\tan\left(u\right)+\sec\left(u\right)\right)}{8}+\dfrac{\sec^3\left(u\right)\tan\left(u\right)}{4}-\dfrac{\sec\left(u\right)\tan\left(u\right)}{8}
\end{equation}
Plugging back the initial change of variable and simplifying we get
\begin{equation}
 {\displaystyle\int}x^2\sqrt{x^2+1}\,\mathrm{d}x
 =\dfrac{\sqrt{x^2+1}\left(2x^3+x\right)-\ln\left(\left|\sqrt{x^2+1}+x\right|\right)}{8}+C
\end{equation}
Now using the integration limits, we have in the simplest form
\begin{equation}
\int_0^1 x^2\sqrt{x^2+1}\,\mathrm{d}x
=
-\dfrac{\operatorname{arsinh}\left(1\right)-3\cdot\sqrt{2}}{8}
\end{equation}
