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Solve the following integral: $$\int \:\frac{1}{\sqrt{x^2-1}}dx$$

I attempted to solve it intergradation by parts by doing a

$$\int \:1\:\frac{1}{\sqrt{x^2-1}} \, dx$$

and set $u$ be $\frac{1}{\sqrt{x^2-1}}$ and $dv/dx$ be $1$:

but as I start doing, it gets more complicated. What is the right direction to solve this?

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up vote 5 down vote accepted

For $x^2-1$ in the radical use $x=\sec\theta$ as $\sec^2\theta-1=\tan^2\theta$

For $x^2+1$ in the radical use $x=\tan\theta$ for the same reason

For $1-x^2$ in the radical use $x=\sin\theta$ as $1-\sin^2\theta=\cos^2\theta$

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I have just found the formalized list : – lab bhattacharjee Oct 21 '13 at 19:12
Substituting $x = \sec \theta$ produces $\int \sec\theta\,d\theta$ if I haven't miscomputed. Does that have an obvious primitive? – Daniel Fischer Oct 21 '13 at 19:16
I know. If it has its own wikipedia article, would you call it obvious? I'd use the $x = \cosh t$ substitution here. – Daniel Fischer Oct 22 '13 at 9:24

Here's an alternative way doing it. Rewrite the integral as:

$\int \frac{1}{\sqrt{x^2-1}} dx = \int \frac{1}{\sqrt{x^2-1}} \cdot \frac{x + \sqrt{x^2-1}}{x + \sqrt{x^2-1}} dx = \int \frac{1 + \frac{x}{\sqrt{x^2-1}} dx}{x + \sqrt{x^2-1}}$.

Now use a simple substitution $u=x+\sqrt{x^2-1}$, so that the integral becomes:

$\int \frac{du}{u} = \log u + C = \log{x+\sqrt{x^2-1}}+C.$

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Substitute $\displaystyle x=\frac1y$ and then $\displaystyle 1-y^2=u^2$

or directly, $\displaystyle x=\frac1y=\frac1{\sqrt{1-u^2}}\implies x^2-1=\frac{u^2}{1-u^2}$

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the solution is $= \operatorname{Argch(x)}$

because $\operatorname{Argch}^{\prime}(x)= \frac{1}{\sqrt{x^2-1}}$.

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Why the $\prime$ in $\operatorname{Argch}^{\prime}(x)$? (I dumbly $\LaTeX$-ed up what you wrote, so left it in.) – user1729 Oct 21 '13 at 19:50
@user1729 That would be the derivative. And I strongly suspect it is meant to be $\operatorname{Ar cosh}(x)$, the area cosinus hyperbolici, since that's the primitive of $\dfrac{1}{\sqrt{x^2-1}}$. – Daniel Fischer Oct 21 '13 at 20:12
@DanielFischer Ah, okay. As I said, I was doing it without thinking. I figured $\operatorname{Argch}$ was something like Arctan on steroids. – user1729 Oct 21 '13 at 20:16

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