# Evaluating $\int \frac{1}{x\sqrt{9x^2-1}}\,dx$

I try to integrate

$$\int \frac{1}{x\sqrt{9x^2-1}}\,dx$$

let $u=x^2,\quad \quad du=2x\,dx,\:\quad \:dx=\frac{1}{2x}\,du$

\begin{align} & \int \frac{1}{x\sqrt{9u-1}}\frac{1}{2x}\,du \\[8pt] = {} & \int \frac{1}{2x^2\sqrt{9u-1}} \, du \\[8pt] = {} & \frac{1}{2}\int \frac{1}{u\sqrt{9u-1}}\, du \end{align}

now let $v=\sqrt{9u-1},\quad \quad dv=\frac{9}{2\sqrt{9u-1}}du,\quad \:dv=\frac{9}{2v}du,\:\quad \:du=\frac{2v}{9}dv$

$$\frac{\int \frac{1}{uv}\frac{2v}{9}dv}{2} = \frac{\int \frac{1}{u}dv}{9}$$

let $v=\sqrt{9u-1},\:u=\frac{v^2+1}{9}$

$$\frac{\int \frac{9}{v^2+1}dv}{9} = \int \frac{dv}{v^2+1}$$

since $\int \frac{1}{v^2+1}\,dv=\arctan(v)$

substitute back $v$ and $u$, then i get

$$\arctan \left(\sqrt{9x^2-1}\right)$$

this is my answer, but i'm not sure if my answer is correct or not. Please if you have a better calculation than mine, i would be really happy if you want to show me and correct my answer. Thank you so much.

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I didn't check your answer but look at this. Your integrand is the derivative of an inverse trig function: coastal.edu/mathcenter/HelpPages/Handouts/invhyp.PDF – TylerHG Sep 1 '14 at 5:10

Hint

I suppose that you could go faster to the solution if you start changing variable $$\sqrt{9 x^2-1}=u$$ that is to say $$x=\frac{\sqrt{u^2+1}}{3}$$ $$dx=\frac{u}{3 \sqrt{u^2+1}} du$$ and then $$\int \frac{dx}{x\sqrt{9x^2-1}}=\int \frac{du}{u^2+1}$$ The whole idea was just to get rid of the radical as fast as possible.

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Apart from the missing $+C$, everything is fine. Well done.

Please note that you can check yourself whether you are right: all you need to do is differentiate.

The "standard" substitution for this kind of problem is $3x=\sec t$.

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thank you, I just did differentiate my answer and the result is exactly the same as the anti-derivative. :) – alethiologist Sep 1 '14 at 5:10
You are welcome. Differentiation is "easy," and can usually be done in one's head. Because I am error-prone, I check automatically. A related little trick is that when I am integrating, I don't worry about multiplicative constants. I forget about such constants, check by differentiating, and adjust the constant at the end. – André Nicolas Sep 1 '14 at 5:13

Looks right as far as I can see, but you always can, and should, check by differentiating your answer and seeing that you get the given integrand.

Slightly quicker method: combine your two substitutions into one by taking $v=\sqrt{9x^2-1}$. But there is not much difference and it is no big deal.

Alternatively, you will find that substituting $$x=\frac{\sec\theta}{3}$$ works beautifully ;-)

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