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



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.

  • 2
    $\begingroup$ Take $x = \sec(\theta)$ $\endgroup$ – Prahlad Vaidyanathan Aug 16 '13 at 1:44
  • $\begingroup$ Shouldn't $du$ be equal to $\frac{2x}{\sqrt{x^2-1}}$? $\endgroup$ – joejacobz Aug 16 '13 at 1:46
  • $\begingroup$ @joejacobz I don't think so derivative-calculator.net/#expr=sqrt%28x%5E2-1%29 $\endgroup$ – Jeel Shah Aug 16 '13 at 1:47
  • 1
    $\begingroup$ @joejacobz Nope, you're missing a $1/2$ factor from the surd. $\endgroup$ – Pedro Tamaroff Aug 16 '13 at 1:48
  • 1
    $\begingroup$ 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. $\endgroup$ – 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, $\displaystyle\Big(x\,dx\Big)$ becomes $\displaystyle\Big( \frac12\,du\Big)$. 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.

  • $\begingroup$ Thanks for showing two ways! $\endgroup$ – Jeel Shah 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.

  • $\begingroup$ 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.) $\endgroup$ – 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: $$\int \frac {du}{u^2 + 1} = \arctan(u) + C = \arctan(\sqrt{x^2 - 1}) + C$$


You very first step should be to make substitution $x=\sec(u)$ instead of $y=\sqrt{x^2-1}$. Then $\sqrt{x^2-1}=\sqrt{\sec^2(u)-1}=\tan(u)$. Also $dx=\sec(u)\tan(u)du$.

And integral becomes ordinary trig integral.

But integral calculator with steps shows another possibility. Start from $y=x^2$, then make another substitution and finally you will arrive to simple rational function.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.