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.

Another way integrate $\sqrt{2 + t^2}$ is to use the Euler substitution $u = \sqrt{2+t^2} + t$. Subtracting $t$ and squaring gives $t =  \frac{1}{2}(u - \frac{2}{u})$. This implies 
$$\sqrt{2+t^2} = u - t = \frac{1}{2}(u + \frac{2}{u})$$
and 
$$dt = \frac{1}{2}(1 + \frac{2}{u^2})\,du$$
Thus 
\begin{align*}
\int \sqrt{2+t^2}\,dt 
&= \int \frac{1}{4}(u + \frac{4}{u} + \frac{4}{u^3})\,du \\
&= \frac{1}{8}u^2 - \frac{1}{2u^2} + \ln{|u|} + C \\
&= \frac{1}{8}(u - \frac{2}{u})(u + \frac{2}{u}) + \ln{|u|} + C \\
&= \frac{1}{2}t\sqrt{2 + t^2} + \ln(t + \sqrt{2 + t^2}) + C
\end{align*}
Applying this antiderivative to your integral, you get $$20\sqrt{402} + \ln(20 + \sqrt{402}) - \ln(-20 + \sqrt{402}) = 2(10\sqrt{402} + \ln(10\sqrt{2} + \sqrt{201}))$$
You can check that $\ln(10\sqrt{2} + \sqrt{201}) = \operatorname{arcsinh}(10\sqrt{2})$ so that this solution is the same as the one in your question.
