<s>[INST] Evaluating $\int_{-20}^{20}\sqrt{2+t^2}\,dt$ I have this integral:
$$\int_{-20}^{20}\sqrt{2+t^2}\,dt$$
I tried solving it many times but without success.
The end result is this:
$$2\left( 10\sqrt{402}+\mathop{\mathrm{arcsinh}}(10\sqrt{2})\right).$$
I can't seem to get this end result. I got a few wrong ones but cant find this one. Perhaps it's wrong? Could anyone confirm it?
I tried Sage, and it calculates it correctly, but not with steps.
 [/INST] While I was writting this answer, Artem posted his. My answer is essentially his 1st one. 
Let $f(t)=\sqrt{2+t^{2}}$. Since $f(-t)=$ $f(t)$, we have
$$\begin{equation*}
I=\int_{-20}^{20}\sqrt{2+t^{2}}dt=2\int_{0}^{20}\sqrt{2+t^{2}}dt
\end{equation*}$$
As the integrand $f(t)$ is a quadratic irrational we can also use the
following Euler substitution
$$\begin{eqnarray*}
x &=&\sqrt{2+t^{2}}-t\Leftrightarrow t=\frac{2-x^{2}}{2x}\Leftrightarrow 
\sqrt{2+t^{2}}=\frac{2+x^{2}}{2x} \\
dt &=&-\frac{2+x^{2}}{2x^{2}}dx,
\end{eqnarray*}$$
which transforms the integrand into a rational fraction in $x$
$$\begin{eqnarray*}
\int \sqrt{2+t^{2}}dt &=&\int \frac{2+x^{2}}{2x}\left( -\frac{2+x^{2}}{2x^{2}
}\right) dx \\
&=&-\int \frac{1}{x}+\frac{1}{x^{3}}+\frac{x}{4}dx \\
&=&-\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{x^{2}}{8}
\right) +C.
\end{eqnarray*}$$
The limits of integration of $I/2$ are
$$\begin{eqnarray*}
t &=&0,x=\sqrt{2} \\
t &=&20,x=\sqrt{2+20^{2}}-20=\sqrt{402}-20.
\end{eqnarray*}$$
Thus
$$\begin{eqnarray*}
I &=&\left. -2\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{
x^{2}}{8}\right) \right\vert _{\sqrt{2}}^{\sqrt{402}-20} \\
&=&2\ln \left( \frac{\sqrt{2}}{\sqrt{402}-20}\right) +\frac{1}{\left( \sqrt{
402}-20\right) ^{2}}-\frac{\left( \sqrt{402}-20\right) ^{2}}{4}+\frac{1}{2}.
\end{eqnarray*}$$
</s>