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.
 A: 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
$$\begin{eqnarray*}
I &=&\left. -2\left( \ln \left\vert x\right\vert -\frac{1}{2x^{2}}+\frac{
x^{2}}{8}\right) \right\vert _{\sqrt{2}}^{\sqrt{402}-20} \\
&=&2\ln \left( \frac{\sqrt{2}}{\sqrt{402}-20}\right) +\frac{1}{\left( \sqrt{
402}-20\right) ^{2}}-\frac{\left( \sqrt{402}-20\right) ^{2}}{4}+\frac{1}{2}.
\end{eqnarray*}$$
A: The typical way to start dealing with an integral that has $\sqrt{a^2+t^2}$ (in your case, $a=\sqrt{2}$) is to use a trigonometric substitution or a hyperbolic substitution.
If we use hyperbolic substitution, we want to use the fact that
$$1 + \sinh^2 z = \cosh^2 z.$$
Set $t=\sqrt{2}\sinh z$. Then
$$\sqrt{2 + t^2} = \sqrt{2 + 2\sinh^2z} = \sqrt{2(1+\sinh^2 z)} = \sqrt{2\cosh^2 z} = \sqrt{2}\cosh z.$$
Also, if $t=\sqrt{2}\sinh z$, then $dt =\sqrt{2}\cosh z\,dz$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\cosh z \sqrt{2}\cosh z\,dz = 2\int\cosh^2 z\,dz.$$
Now we need to solve the integral of $\cosh^2 z$. We can do this by parts, similarly to how we solve the integral of $\cos^2t$. Let $u=\cosh z$, $dv=\cosh z\,dz$. Then $du=\sinh z\,dz$, $v=\sinh z$, so
$$\begin{align*}
\int\cosh^2 z\,dz &= \cosh z\sinh z - \int \sinh^2z\,dz\\
&= \cosh z \sinh z - \int(cosh^2z - 1)\,dz\\
&= \cosh z\sinh z +z - \int\cosh^2z\,dz.
\end{align*}$$
Hence
$$\begin{align*}
\int\cosh^2z\,dz &= z + \cosh z\sinh z - \int\cosh^2z\,dz\\
2\int\cosh^2z\,dz &= z+ \cosh z\sinh z + C\\
\int\cosh^2 z\,dz &= \frac{1}{2}z + \frac{1}{2}\cosh z\sinh z + C.
\end{align*}$$
Then to get it back into a function of $t$ we remember that $t=\sqrt{2}\sinh z$, so $\sinh z = \frac{\sqrt{2}}{2}t$. Then $z= \mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right)$, and
$$\sinh z\cosh z = \sinh z \sqrt{1+\sinh^2z} = \frac{\sqrt{2}}{2}t\sqrt{1 + \frac{1}{2}t^2} = \frac{1}{2}\sqrt{2+t^2}.$$
Therefore,
$$\int\sqrt{2+t^2}\,dt = 2\int\cosh^2z\,dz = 2\mathop{\mathrm{arcsinh}}\left(\frac{\sqrt{2}}{2}t\right) + \sqrt{2+t^2}+C.$$
Now plugging into the definite integral gives you a solution.

If we use trigonometric substitutions, we want to use the fact that
$$1 + \tan^2 \theta = \sec^2\theta$$
Set $t=\sqrt{2}\tan\theta$, with $-\frac{\pi}{2} \lt \theta\lt\frac{\pi}{2}$. Then
$$\sqrt{2+t^2} = \sqrt{2+2\tan^2\theta} = \sqrt{2}\sqrt{1+\tan^2\theta} = \sqrt{2}\sqrt{\sec^2\theta} = \sqrt{2}|\sec\theta|.$$
Since $\sec\theta\gt 0$ on $-\frac{\pi}{2}\lt \theta\lt \frac{\pi}{2}$, we get that $\sqrt{2+t^2} = \sqrt{2}\sec\theta$.
Also, if $t=\sqrt{2}\tan\theta$, then $dt = \sqrt{2}\sec^2\theta\,d\theta$. Therefore,
$$\int\sqrt{2+t^2}\,dt = \int \sqrt{2}\sec\theta\sqrt{2}\sec^2\theta\,d\theta = 2\int\sec^3\theta\,d\theta.$$
So we need to find $\int\sec^3\theta\,d\theta = \int\frac{1}{\cos^3\theta}\,d\theta$.  
This can be done any number of ways. Using integration by parts, we get
$$\int\frac{d\theta}{\cos^3\theta} = \frac{1}{2}\tan\theta + \frac{1}{2}\int\frac{d\theta}{\cos\theta}.$$
And
$$\int\frac{d\theta}{\cos\theta} = \int\sec\theta\,d\theta = \ln|\sec\theta + \tan\theta|+C.$$
Thus, we get that
$$\int\sqrt{2+t^2}\,dt = \tan\theta + \ln|\sec\theta+\tan\theta|+C.$$
To change it into a formula using $t$, we use the fact that $t=\sqrt{2}\tan\theta$. Therefore, $\tan\theta = \frac{\sqrt{2}}{2}t$; and 
$$\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + \frac{t^2}{2}} = \frac{\sqrt{2}}{2}\sqrt{2+t^2}.$$
Plugging in and evaluating gives the desired result.
A: Here are two more ways to calculate $\int \sqrt{a^2+x^2}\ dx$.


*

*Let us make the substitution $\sqrt{a^2+x^2}=t-x$. After squaring, $a^2=t^2-2xt$, from which
$$
x=\frac{t^2-a^2}{2t}\,.
$$ 
Using the last expression, one finds
$$
\sqrt{a^2+x^2}=t-x=\frac{t^2+a^2}{2t}\,,
$$
and
$$
dx=\frac{t^2+a^2}{2t^2}dt.
$$
Plugging everything into the integral,
$$
\frac 14\int\left(\frac{2a^2}{t}+t+\frac{a^4}{t^3}\right)dt=\frac{a^2}{2}\ln t+\frac{t^4-a^4}{8t^2}+C.
$$
Note, that from the above $x\sqrt{a^2+x^2}=\frac{t^4-a^4}{4t^2}$, therefore we obtain finally
$$
\int\sqrt{a^2+x^2}\,dx=\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+\frac{1}{2}x\sqrt{a^2+x^2}+C.
$$

*Consider the integration by parts:
$$
u=\sqrt{a^2+x^2},\quad v=x,\quad du=\frac{x}{\sqrt{x^2+a^2}}.
$$
Therefore,
$$
\int\sqrt{x^2+a^2}\,dx=x\sqrt{a^2+x^2}-\int\frac{x^2dx}{\sqrt{a^2+x^2}}=\\
x\sqrt{a^2+x^2}-\int\frac{x^2+a^2-a^2 dx}{\sqrt{a^2+x^2}}=\\x\sqrt{a^2+x^2}-\int\sqrt{a^2+x^2}\,dx+\int\frac{a^2dx}{\sqrt{a^2+x^2}}.
$$
It is known (and can easily be found with the variable change from my point 1) that
$$
\int\frac{dx}{\sqrt{a^2+x^2}}=\ln(x+\sqrt{a^2+x^2})+C,
$$
hence, finally,
$$
\int\sqrt{a^2+x^2}\,dx=\frac{a^2}{2}\ln(x+\sqrt{a^2+x^2})+\frac{1}{2}x\sqrt{a^2+x^2}+C.
$$

A: 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.
