Integration question: $\int \frac{\mathrm{d}x}{\sqrt{3 x} (3 x+1)}$

Question

I am missing one piece of how to integrate the following:

$\int \frac{\mathrm{d}x}{\sqrt{3 x} (3 x+1)}$

I found a solution to a similar problem which I entirely understand:

I can usually use Wolfram Alpha's step-by-step solutions to clarify if I'm having trouble understanding a problem but sometimes it doesn't use the most efficient and/or understandable steps and it's not much help in this case:

The part that I'm not understanding is how the 3 ends up in the denominator of the leading coefficient as in the step circled above. My goal here is not to just get the answer (which I obviously already have) but rather to completely understand how to solve this and similar problems. I'm also not necessarily interested in an explanation of the Wolfram steps but rather the most logical method(s) for solving it.

Thanks for any help.

P.S. I apologize in advance if this is a duplicate. I looked through a ton of problems using search and didn't see anything that addressed this specific type of problem.

Answer 1

Note that $u=\frac{s}{\sqrt{3}}$ and $u^2=\frac{s^2}{3}$ and that $3u^2 = s^2$. Further note that $ds=\frac{du}{\sqrt{3}}$

So after making the substitution, what you have is:

$$ \frac{2}{\sqrt{3}} \int\frac{1}{s^2+1}\frac{ds}{\sqrt{3}} $$

Does it make more sense now?

Answer 2

It is almost exactly the problem you understand. Let $3x=t$. Then $dx=\frac{1}{3}\,dt$, and $$\int \frac{dx}{\sqrt{3x}(3x+1)}=\int \frac{1}{3}\frac{dt}{\sqrt{t}(t+1)}.$$

Answer 3

I'd suggest you start with $$u^2 = 3x \implies 2u\,du = 3\,dx \iff \frac 23\,du = \frac{dx}{u} = \frac{dx}{\sqrt{3x}}$$

Then you have $$\int \frac{dx}{\sqrt {3x}}\cdot \frac 1{(3x+1)} = \frac 23 \int \frac{du}{u^2 + 1}$$

Now you can put $u = \tan \theta$, and proceed from there.