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

Question

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.

Answer 1

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

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.

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

$$ \begin{aligned}\int \frac{d x}{x \sqrt{x^{2}-1}} =\int \frac{1}{x^{2}} d\left(\sqrt{x^{2}-1}\right) =\int \frac{d\left(\sqrt{x^{2}-1}\right)}{\left(\sqrt{x^{2}-1}\right)^{2}+1} =\tan ^{-1}\left(\sqrt{x^{2}-1}\right)+C \end{aligned} $$

Answer 6

$$u^{2}=x^{2}-1\Rightarrow xdx=udu\\ \int \frac{dx}{x\sqrt{x^2-1}}=\int \frac{xdx}{(x^{2}-1+1)\sqrt{x^2-1}}\\ =\int\frac{udu}{(u^{2}+1)u} =\int\frac{du}{u^{2}+1}=\arctan(u)+c=\arctan\sqrt{x^{2}-1}+c$$