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by **stellar** on Sun May 17, 2009 2:55 am

How do I integrate [1/(2e^x - 3)]dx using substitution?

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by **Martingale** on Sun May 17, 2009 4:02 am

stellar wrote:How do I integrate [1/(2e^x - 3)]dx using substitution?

let $u=2e^x-3$

then $u+3=2e^x$

and $du=2e^xdx\Rightarrow du=(u+3)dx\Rightarrow \frac{du}{u+3}=dx$

then the integral becomes

$\int \frac{1}{u(u+3)}du$

then use partial fractions

by **stellar** on Mon May 18, 2009 8:15 pm

i dont know why i couldnt see that earlier
thanks!

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How do I integrate [1/(2e^x - 3)]dx using substitution?

How do I integrate [1/(2e^x - 3)]dx using substitution?

Re: How do I integrate [1/(2e^x - 3)]dx using substitution?

Re: How do I integrate [1/(2e^x - 3)]dx using substitution?