Difficult integration

**Chicken Enchilada** · September 13th 2009, 07:48 PM

I'm having some trouble with a more complicated problem.

$\int \frac {x^{3x}} {\sqrt {4 - e^{6x}}}$

So far I have:

$u = e ^{3x}$ so $\frac {1}{3}du = e^{3x}dx$

That gives me:

$\frac {1}{3}\int \frac {x^{3x}}{\sqrt{4-u^2}}dx$

I need help getting rid of that x on top and turn that 4 into a 1. I'm sure that I'm missing something obvious.

**mr fantastic** · September 14th 2009, 02:43 AM

Originally Posted by Chicken Enchilada

I'm having some trouble with a more complicated problem.

$\int \frac {x^{3x}} {\sqrt {4 - e^{6x}}}$

So far I have:

$u = e ^{3x}$ so $\frac {1}{3}du = e^{3x}dx$

That gives me:

$\frac {1}{3}\int \frac {x^{3x}}{\sqrt{4-u^2}}dx$

I need help getting rid of that x on top and turn that 4 into a 1. I'm sure that I'm missing something obvious.

I suspect the question is meant to be $\int \frac {{\color{red}e}^{3x}} {\sqrt {4 - e^{6x}}} \, {\color{red}dx}$

**Chicken Enchilada** · September 14th 2009, 02:49 PM

Originally Posted by mr fantastic

I suspect the question is meant to be $\int \frac {{\color{red}e}^{3x}} {\sqrt {4 - e^{6x}}} \, {\color{red}dx}$

Good point. I will take it for a typo then. I got all the other problems done, so I think I'm set. Thank You.

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