1. ## Integral Help

How do I begin to solve this?

2. by first making the substitution $t=\sqrt{3x-4}.$

3. Originally Posted by Krizalid
by first making the substitution $t=\sqrt{3x-4}.$
That helped, but I'm still having trouble. What's after the substitution?

4. note that $dt=\frac{3}{2\sqrt{3x-4}}\,dx,$ and $x=\frac{t^2+4}3,$ can you take it from there now?

5. Originally Posted by Krizalid
note that $dt=\frac{3}{2\sqrt{3x-4}}\,dx,$ and $x=\frac{t^2+4}3,$ can you take it from there now?
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?

6. $\int{\frac{dx}{x\sqrt{3x-4}}},$ put $t=\sqrt{3x-4}$ then $\frac{2}{3}\,dt=\frac{dx}{\sqrt{3x-4}},$ so $\int{\frac{dx}{x\sqrt{3x-4}}}=2\int{\frac{dt}{t^{2}+4}\,dt}=\arctan \left( \frac{t}{2} \right)+k.$

7. Originally Posted by Krizalid
$\int{\frac{dx}{x\sqrt{3x-4}}},$ put $t=\sqrt{3x-4}$ then $\frac{2}{3}\,dt=\frac{dx}{\sqrt{3x-4}},$ so $\int{\frac{dx}{x\sqrt{3x-4}}}=2\int{\frac{dt}{t^{2}+4}\,dt}=\arctan \left( \frac{t}{2} \right)+k.$
Thank you, I just made this a lot harder than it actually was.