How to calculate $\int_{0}^{1} x^2 \sqrt{1+x^2} dx$?

Question

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.

Answer 1

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:

$$\int_0^{\operatorname{Arsinh 1}} \sinh^2u\cosh^2u\,du = \frac1{16} \int_0^{\operatorname{Arsinh 1}} (e^{2u}+e^{-2u} - 2) (e^{2u}+e^{-2u} + 2)\,du$$

Answer 2

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}