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How do I begin to solve this?

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- Apr 3rd 2010, 01:27 PMSimon777Integral Help
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How do I begin to solve this? - Apr 3rd 2010, 02:17 PMKrizalid
by first making the substitution $\displaystyle t=\sqrt{3x-4}.$

- Apr 3rd 2010, 02:28 PMSimon777
- Apr 3rd 2010, 02:32 PMKrizalid
note that $\displaystyle dt=\frac{3}{2\sqrt{3x-4}}\,dx,$ and $\displaystyle x=\frac{t^2+4}3,$ can you take it from there now?

- Apr 3rd 2010, 02:56 PMSimon777
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 PMKrizalid
$\displaystyle \int{\frac{dx}{x\sqrt{3x-4}}},$ put $\displaystyle t=\sqrt{3x-4}$ then $\displaystyle \frac{2}{3}\,dt=\frac{dx}{\sqrt{3x-4}},$ so $\displaystyle \int{\frac{dx}{x\sqrt{3x-4}}}=2\int{\frac{dt}{t^{2}+4}\,dt}=\arctan \left( \frac{t}{2} \right)+k.$

- Apr 3rd 2010, 03:34 PMSimon777