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!!! =)