Common Integral? So I came across the integral of $$\frac{1}{\sqrt{x^2+1}}$$ and I've tried substitution and it isn't on the list of common integrals. I used Wolfram Alpha, but it said $\sinh x$, but I was wondering if there is an alternate answer?
 A: Two substitutions suggest themselves
$x=\sinh(u)\Rightarrow\sqrt{1+x^2}=\cosh(u)\text{ and }\mathrm{d}x=\cosh(u)\,\mathrm{d}u$ which yields
$$
\begin{align}
\int\frac{\mathrm{d}x}{\sqrt{1+x^2}}
&=\int\frac{\cosh(u)\,\mathrm{d}u}{\cosh(u)}\\
&=\int\,\mathrm{d}u
\end{align}
$$
$x=\tan(v)\Rightarrow\sqrt{1+x^2}=\sec(v)\text{ and }\mathrm{d}x=\sec^2(v)\,\mathrm{d}v$ which yields
$$
\begin{align}
\int\frac{\mathrm{d}x}{\sqrt{1+x^2}}
&=\int\frac{\sec^2(v)\,\mathrm{d}v}{\sec(v)}\\
&=\int\frac{\mathrm{d}\sin(v)}{1-\sin^2(v)}\\
&=\frac12\int\frac{\mathrm{d}\sin(v)}{1-\sin(v)}+\frac12\int\frac{\mathrm{d}\sin(v)}{1+\sin(v)}
\end{align}
$$
A: You can some times find antiderivatives by looking in lists of derivatives. See for example here: http://en.wikipedia.org/wiki/Differentiation_rules#Derivatives_of_hyperbolic_functions
So you have
$$
\int \frac{1}{\sqrt{1+x^2}}\; dx = \text{arsinh}(x) + C = \sinh^{-1}(x) + C.
$$
A: It should be $\sinh^{-1} x$, which can also be written as $\ln (1+\sqrt{1+x^2})$ as shown in Wikipedia.
A: The way to do it is via trigonometric substitution. If we substitute $x=\tan\theta$, then $\frac{1}{\sqrt{x^2+1}} =\frac{1}{\sqrt{1+\tan^2\theta}} = \frac{1}{\sec\theta}$ and $dx=\sec^2\theta d\theta$, so the integral becomes
$\displaystyle \int \frac{1}{\sec\theta}\sec^2\theta d\theta = \int \sec\theta d\theta$, with appropriately changed limits of integration. The indefinite integral turns out to be $\log|\sec \theta + \tan\theta|+C$.
Now using the original substitution $x=\tan\theta$ it is possible to express $\sec\theta$ and $\tan \theta$ in terms of $x$.
