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Mathematics 17 Online
OpenStudy (luigi0210):

What could I use as a u-sub?

OpenStudy (luigi0210):

\[\LARGE \int~\frac{e^x}{3-e^{2x}}~dx\]

OpenStudy (muzzack):

The method of u-substitution is a method for algebraically simplifying the form of a function so that its antiderivative can be easily recognized. This method is intimately related to the chain rule for differentiation. For example, since the derivative of \[e ^{x}\] is \[D(e ^{x}) = e ^{x}\] it follows easily that \[\int\limits(e ^{x})dx=e ^{x}+C\]

OpenStudy (muzzack):

@mathmale

OpenStudy (anonymous):

\[u=3-e^{2x}~~\Rightarrow~~du=-2e^{2x}~dx~~\iff~~-\frac{1}{2}du=e^{2x}~dx\] \[\int\frac{e^{2x}}{3-e^{2x}}~dx=-\frac{1}{2}\int\frac{du}{u}\]

OpenStudy (anonymous):

Oh sorry, I thought there was an \(e^{2x}\) in the numerator...

OpenStudy (luigi0210):

Yea, that's my problem.

OpenStudy (anonymous):

A trig sub should work: \(e^x=\sqrt 3\sin u\), so that \(e^x~dx=\sqrt3\cos u~du~~\iff~~dx=\dfrac{\sqrt3\cos u}{\sqrt3\sin u}~du=\cot u~du\)

OpenStudy (luigi0210):

Alright, thank you siths and muzzack :)

OpenStudy (anonymous):

yw

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