How can I solve $\int \sqrt{1+x^2}\,dx$? $$\int \sqrt{1+x^2}\,dx$$
I've tried some step-by-step calculators, but what they give is way beyond my level. Is there any "simple" way to solve this?
 A: Several people have pointed out that you can write $x=\tan\theta.$ That gets you $dx=\sec^2\theta\,d\theta$ and $\sqrt{1+x^2}=\sec\theta,$ so you have
$$
\int\sec^3\theta\,d\theta.
$$
This is one of the standard examples of a standard trick:
\begin{align}
\int \sec^3\theta\,d\theta & = \int(\sec\theta) \Big(\sec^2\theta\,d\theta\Big) = \overbrace{\int u\,dv = uv -\int v\,du}^{\Large\text{integration by parts}} \\[10pt]
& = \sec\theta\tan\theta - \int \tan^2\theta\sec\theta\,d\theta \\[10pt]
& = \sec\theta\tan\theta - \int(\sec^2\theta-1)\sec\theta\,d\theta \\[10pt]
& = \sec\theta\tan\theta - \int\sec^3\theta\,d\theta + \int\sec\theta\,d\theta. \\[15pt]
\text{Thus we have } & \int \sec^3\theta\,d\theta = \Big(\text{something}\Big) - \int\sec^3\theta\,d\theta. \\[10pt]
\text{Adding the integral to } \\
\qquad \text{both sides, we get } & 2\int\sec^3\theta\,d\theta = \Big(\text{something}\Big). \\[10pt]
\text{And so } & \phantom{2}\int\sec^3\theta\,d\theta = \frac 1 2 \Big(\text{something}\Big) \\[10pt]
= {} & \frac 1 2 \left( \sec\theta\tan\theta + \int \sec\theta\,d\theta\right).
\end{align}
Then you have the problem of finding that last integral, which is somewhat challenging as well.
A: Hint:
try substitute 
$$x=\tan { \theta  } $$
A: $x=\frac{e^t-e^{-t}}{2}$  also helps.
A: This is just straight up trigonometric substitution with $x=\tan(\theta)$ and $dx=\sec^{2}(\theta)\:d\theta$. Then use the identity $1+\tan^{2}(\theta)=\sec^{2}(\theta)$ to eliminate the square root and integrate from there. 
