Indefinite integral So I'm currently trying to solve
$$\int \sqrt{ 1+\frac{1}{3x} }  \, \, dx$$
I know that this can also be represented as ((x+1/3)/x)^1/2 but I dont like that form. I also know that this can be done with sustitution. I've done lot's of stuff but I get stuck everytime. You don't need to solve me the problem, just point me to the right direction if you want to and I'll finish it myself. 
I'll write my first impression. I proceed to choose  $u = x + \frac{1}{3x^{2}}$
so     $du =1 -1/3x^{2} dx$
So I end with $\int \sqrt{u} \, \, \, du -3x^{2} $
I'm sure this is wrong but I don't know why. Maybe it wasn't wise to choose u as the whole square root? I did so because the immediate integral of x^1/2 is easy. Did I do something wrong? Or can I continue from here? If so, how?
Thanks a ton!!! =)
 A: $$\int \sqrt{ 1+\frac{1}{3x} }  \, \, dx$$
let $u^2=3x \implies 2u\ du = 3\ dx$
$$\frac{2}{3}\int \sqrt{ 1+\frac{1}{u^2} }  \, \, u\ du$$
$$\frac{2}{3}\int \sqrt{ u^2+1 }  \, \, du$$
now you can use $u = \sinh(t)$
A: I suggest you to set $u = 3x$, but prior to that, do some thing with the sum under the root. Then use a pretty elementary integration technique.
A: As suggested by busman, it may be a good idea to let $u=3x$. The constants one has to drag around are then less annoying. 
So, apart from a constant factor, we want 
$$\int\sqrt{1+\frac{1}{u}}\,du.$$
Now I would suggest letting $w^2=1+\dfrac{1}{u}$. Then 
$$2w\,dw=-\frac{du}{u^2}=-(w^2-1)^2 \,du.$$
We end up having to find something like
$$\int-\frac{2w^2\,dw}{(w^2-1)^2}.$$
This is a not completely pleasant partial fractions problem. 
A: Well, the problem appears to be the radical.  I'm not sure setting $u$ equal to everything under the radical looks promising, as you still need a $du$.  There is, however, more than one way to get rid of a radical.
If you are comfortable with hyperbolic functions, letting $\frac1{3x}=\sinh^2 u,x=\frac13\operatorname{csch}^2u$.  Myself, I was never taught those and would go with the similar trig substitution $\frac1{3x}=\tan^2u,x=\frac13\cot^2u,dx=-\frac23\cot u\csc^2udu$.
