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Math Help - Integral Help

  1. #1
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    Integral Help




    How do I begin to solve this?
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  2. #2
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    by first making the substitution t=\sqrt{3x-4}.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    by first making the substitution t=\sqrt{3x-4}.
    That helped, but I'm still having trouble. What's after the substitution?
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  4. #4
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    note that dt=\frac{3}{2\sqrt{3x-4}}\,dx, and x=\frac{t^2+4}3, can you take it from there now?
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    Quote Originally Posted by Krizalid View Post
    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?
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  6. #6
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    \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.
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  7. #7
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    Quote Originally Posted by Krizalid View Post
    \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.
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