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how do I solve the integral 2/(e^(2x)+4) ? using the substitution u=e^(2x)+4
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\[\int\frac{2}{e^{2x}+4}dx\] \[u=e^x\\du=e^x\;dx\\ \frac{1}{e^x}du=dx\\ \frac{1}{u}du=dx\] \[2\int\frac{1}{u^2+4}\left(\frac{1}{u}du\right)\\ 2\int\frac{1}{u(u^2+4)}du\] Partial fraction decomposition?
\[\Large \frac{2}{e^{2x}+4}\cdot\frac{e^{-2x}}{e^{-2x}}=\frac{2e^{-2x}}{1+4e^{-2x}}\]
Oh, sorry. Didn't notice the substitution requirement.
it can also be solved by partial fraction.
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