Integrating $\int\frac{2dx}{x\ln(6x)}$

I needed some help integrating this: $$\int\frac{2\,dx}{x\ln(6x)}.$$

I have never seen the dx within the problem like that, I am assuming I can't just move it to the outside can I?

Can I start by factoring out the 2 and then setting $u = \ln(6x)$ and $du = (1/x)\; dx$.

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Move the dx to the outside of what? It is a part of the integrand just as it should be... – Brandon Carter Jan 31 '11 at 4:08
@Finzz: It's a bit of abuse of notation, based on the fact that $\frac{a}{b}\times c = \frac{ac}{b}$. So $$\int\frac{2dx}{x\ln(6x)} = \int\left(\frac{2}{x\ln(6x)}\right)\;dx.$$ As to your $u$ and $du$, you got $du$ slightly wrong (you forgot the $dx$). – Arturo Magidin Jan 31 '11 at 4:09
@Brandon @ Arturo: Alright, then that's what I thought. I have to enter my answers into an online homework system that determines if your answer is correct or not. I worked it out to be 2(log(log6x)) and it says it is incorrect so I came on here for help. I guess I will email my instructor about it. – Finzz Jan 31 '11 at 4:10
@Finzz: For starters, you forgot the constant of integration. For seconds, you forgot the absolute value ($\int\frac{1}{u}\,du = \ln|u|+C$). And since the problem uses ln instead of log, I'll wager the answer system expects ln, not log. That may be enough to say you have the wrong answer... I certainly take points off for the former two mistakes when I grade. – Arturo Magidin Jan 31 '11 at 4:13
@Arturo Sorry, I forgot to mention that the answer system uses log for ln and log10 for log. I tried it both ways and still got it wrong. – Finzz Jan 31 '11 at 4:22

Yes, you can. In this, $dx$ is a term you can move wherever following the commutativity and associativity of multiplication. Your approach is quite reasonable.

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You have the right idea by pulling out any constants and letting the natural logarithm become u. So we are trying to solve the following integral. $$\int\frac{2\,dx}{x\ln(6x)}.$$

When looking at ugly integrands as such, we try to use the most simple methods that are allowed to compute the integral. So for our case, the simplest would be to try u-substitution first. So by doing this, it leads:

u = ln(6x)

du = $\dfrac{6}{6x}dx = \dfrac{1}{x}dx$

dx = x $du$

To brush up on our derivatives for those not seeing how we got that, it follows from first semester calculus or Calculus I that $\dfrac{d}{dx}$ $\Bigg[\text{ln}\Big(f(x)\Big)\Bigg]~=~\dfrac{\dfrac{d}{dx}f(x)}{f(x)}$.

So with this being known now, we can now substitute in our u and dx into the integral leading us to: $$\int\frac{2}{xu}\cdot x~du$$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~~~~~$ $\displaystyle\int\frac{2}{u}\cdot du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~$ $\displaystyle2\int\frac{1}{u}\cdot du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~~~~~2\cdot \text{ln}|u|+C$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\therefore~~~~~2\cdot \text{ln}\Big|\text{ln(6x)}\Big|+C$

We have now found what the integral evaluates to.

Hope that this helped. Let me know if there is anything step that you would like for me clear up and further clarify why I did so.

Thanks

-Good~Luck

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Here is an other useful approach. By the chain rule $$f(g(x))'= f'(g(x))g'(x)$$ and hence, $$\frac{d}{dx}\log(f(x))=\frac{f'(x)}{f(x)}$$ (this is called the logarithmic derivative of $f$). In your case put $f(x)=\log(6x)$ to get $$\frac{d}{dx}\log(f(x))=\frac{f'(x)}{f(x)}=\frac{1/x}{\log(6x)}=\frac{1}{x\log(6x)}.$$

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