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?

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Shouldn't it be sinh^{-1}x? – Amzoti Dec 12 '12 at 3:31

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}

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It should be $\sinh^{-1} x$, which can also be written as $\ln (1+\sqrt{1+x^2})$ as shown in Wikipedia.

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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 $$\int \frac{1}{\sqrt{1+x^2}}\; dx = \text{arsinh}(x) + C = \sinh^{-1}(x) + C.$$

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Its the square root of 1+x^2. – Lizi Dec 12 '12 at 3:28
Missed the sqrt in the denominator. – Amzoti Dec 12 '12 at 3:28

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 $x$.

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