# Calculate $\int \frac{dx}{x\sqrt{x^2-1}}$

I am trying to solve the following integral

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

I did the following steps by letting $u = \sqrt{x^2-1}$ so $\text{d}u = \dfrac{x}{\sqrt{{x}^{2}-1}}$ then

\begin{align} &\int \frac{\sqrt{x^2-1} \, \text{d}u}{x \sqrt{x^2-1}} \\ &\int \frac{1}{x} \text{d}u \\ &\int \frac{1}{\sqrt{u^2+1}} \text{d}u\\ \end{align}

Now, this is where I am having trouble. How can I evaluate that? Please provide only hints

Thanks!

EDIT:

The problem specifically states that one must use substitution with $u = \sqrt{x^2-1}$. This problem is from the coursera course for Single Variable Calculus.

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Take $x = \sec(\theta)$ –  Prahlad Vaidyanathan Aug 16 '13 at 1:44
Shouldn't $du$ be equal to $\frac{2x}{\sqrt{x^2-1}}$? –  joejacobz Aug 16 '13 at 1:46
@joejacobz I don't think so derivative-calculator.net/#expr=sqrt%28x%5E2-1%29 –  gekkostate Aug 16 '13 at 1:47
@joejacobz Nope, you're missing a $1/2$ factor from the surd. –  Pedro Tamaroff Aug 16 '13 at 1:48
Four answers are seen below (if you don't count the deleted one), and yet I'm the only one who's up-voted the question so far (unless maybe an up-vote and a down-vote canceled each other?). Despite all the answers already here, I posted one of my own, with two different methods, and at least some hints---or maybe slightly more than just hints--- about recognizing when to use certain methods. –  Michael Hardy Aug 16 '13 at 2:56

$$\int\frac{dx}{x\sqrt{x^2+1}} = \int\frac{x\,dx}{x^2\sqrt{x^2+1}} = \int\frac{1}{x^2\sqrt{x^2+1}} \Big(x\,dx\Big)$$

The big parentheses are of course a hint that what's inside them is to become $du$, or a constant times $du$. But should $u$ be $x^2$ or $x^2+1$? Either way, $\displaystyle\Big(x\,dx\Big)$ becomse $\displaystyle\Big( \frac12\,du\Big)$. I think usually it's better to have the thing under the radical be simple, so I'll say $u=x^2+1$, and we have $$\frac12\int\frac{du}{(u-1)\sqrt{u}}.$$ We can rationalize $\sqrt{u}$ by letting \begin{align} w & = \sqrt{u} \\ w^2 & = u \\ 2w\,dw & = du \end{align} and we have $$\frac12\int\frac{2w\,dw}{(w^2-1)w} = \int\frac{dw}{w^2-1}.$$ Then use partial fractions, getting $$\int\left(\frac{A}{w-1}+\frac{B}{w+1}\right)\,dw$$ and you need to figure out what $A$ and $B$ are.

That works, but a trigonometric substitution also comes to mind. The expression $\sqrt{x^2+1}$ should remind you of $\sqrt{\tan^2\theta+1}= \pm\sec\theta$, and if it doesn't remind you of that, that's something to work on. Review some trigonometry and trigonometric substitutions. If $x=\tan\theta$ then $dx=\sec^2\theta\,d\theta$, and we have $$\int\frac{\sec^2\theta\,d\theta}{\tan\theta\sec\theta} = \int\frac{\sec\theta\,d\theta}{\tan\theta} = \int\csc\theta\,d\theta.$$ That's a hard one to do from scratch, but it's also one that you can look up in standard tables.

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Thanks for showing two ways! –  gekkostate Aug 16 '13 at 5:03

You were basically there, just a little slip in the substitution process, you should have ended up with $\frac{1}{u^2+1}$.

Rewrite our integral as $$\int \frac{x\,dx}{x^2\sqrt{x^2-1}}.$$ Make the substitution $u=\sqrt{x^2-1}$. Then $du=\frac{x}{\sqrt{x^2-1}}\,dx$, so $x\,dx=u\,du$.

The rest I leave to you. It will be very easy, one short line.

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Or: $u = \sqrt{x^2-1} \Rightarrow x=\sqrt{u^2=1}$,$dx= u\\ du/\sqrt{u^2+1}$. (I can't get this to format right.) –  Stephen Herschkorn Aug 16 '13 at 2:29

You had the "gist" of what you needed to do, but as others have noted, your substitution should yield the integrand $\dfrac{1}{u^2+1}$.

We have $$\int \frac {dx}{x \sqrt{x^2 - 1}} = \int \frac{x\,dx}{x^2\sqrt{x^2-1}}$$ As you did, we let $\, u=\sqrt{x^2-1}$. Then $du=\frac{x}{\sqrt{x^2-1}}\,dx$, so $x\,dx=\sqrt{x^2 - 1}\,du = u \,du$.

Note that $$u = \sqrt{x^2 - 1} \implies u^2 = x^2 - 1 \iff x^2 = u^2 + 1$$

So substituting gives us $$\int \frac{x\,dx}{x^2\sqrt{x^2-1}} = \int \dfrac{u \,du}{(u^2 + 1)u} = \int \frac {du}{u^2 + 1}$$

Now, we can use trigonometric substitution, and given a denominator of the form $u^2 + 1$, put $u = \tan \theta$. This gives us: $$\int \frac {du}{u^2 + 1} = \arctan(u) + C = \arctan(\sqrt{x^2 - 1}) + C$$

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When $x =\sin u$, $\displaystyle\int\frac{dx}{x\sqrt{x^2 -1}} \, dx$ becomes $\displaystyle\int\frac{\cos u \, du}{\sin u\sqrt{-\cos^2u}}$.

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