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

Question

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 $du = (1/x)\; dx$.

Answer 1

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.

Answer 2

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 $$\frac{d}{dx}\log(f(x))=\frac{f'(x)}{f(x)}=\frac{1/x}{\log(6x)}=\frac{1}{x\log(6x)}.$$

Answer 3

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