# Integral Help

• Apr 3rd 2010, 01:27 PM
Simon777
Integral Help
• Apr 3rd 2010, 02:17 PM
Krizalid
by first making the substitution $t=\sqrt{3x-4}.$
• Apr 3rd 2010, 02:28 PM
Simon777
Quote:

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?
• Apr 3rd 2010, 02:32 PM
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?
• Apr 3rd 2010, 02:56 PM
Simon777
Quote:

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?
• Apr 3rd 2010, 03:16 PM
Krizalid
$\int{\frac{dx}{x\sqrt{3x-4}}},$ put $t=\sqrt{3x-4}$ then $\frac{2}{3}\,dt=\frac{dx}{\sqrt{3x-4}},$ so $\int{\frac{dx}{x\sqrt{3x-4}}}=2\int{\frac{dt}{t^{2}+4}\,dt}=\arctan \left( \frac{t}{2} \right)+k.$
• Apr 3rd 2010, 03:34 PM
Simon777
Quote:

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.