How do I begin to solve this?
That got me further along in the problem, but I'm at 3/4 times the integral of cot which I know can't give me the answer to the problem. I've been on this problem for 2 hours and its very frustrating to keep hitting these dead ends. I hate to ask for anymore help from you, but would you mind working it out a bit further?
$\displaystyle \int{\frac{dx}{x\sqrt{3x-4}}},$ put $\displaystyle t=\sqrt{3x-4}$ then $\displaystyle \frac{2}{3}\,dt=\frac{dx}{\sqrt{3x-4}},$ so $\displaystyle \int{\frac{dx}{x\sqrt{3x-4}}}=2\int{\frac{dt}{t^{2}+4}\,dt}=\arctan \left( \frac{t}{2} \right)+k.$