Wikipedia:Reference desk/Archives/Mathematics/2016 July 16

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July 16

Help with Integral

Hi all, how to solve the integral $\int {\frac {1}{e^{2x}+e^{3x}}}dx$? 31.154.81.56 (talk) 07:51, 16 July 2016 (UTC)

The problem is the term $e^{x}$ so try using the substitution $x=\log y$ to simplify. Dmcq (talk) 08:01, 16 July 2016 (UTC)

How should I use it, if there's no multiplication by the derivative ${\frac {1}{y}}$? 31.154.81.56 (talk) 13:24, 16 July 2016 (UTC)

I don't understand your question, there is a multiplication by 1/y at one stage, what problem do you have with that? Dmcq (talk) 13:36, 16 July 2016 (UTC)

As far as I know $\int {\frac {1}{e^{2x}+e^{3x}}}\cdot {\frac {1}{x}}dx=\int {\frac {1}{y^{2}+y^{3}}}dy$, but this is not the case here. Am I wrong? Or maybe you meant something else? 31.154.81.56 (talk) 13:45, 16 July 2016 (UTC)

You forgot the part $dx={\frac {dy}{y}}$ and hence you end up with $\int {\frac {dy}{y^{3}+y^{4}}}$ instead. Factor the denominator to get ${\frac {dy}{y^{3}(y+1)}}$ and apply a partial fraction decomposition.--Jasper Deng (talk) 15:25, 16 July 2016 (UTC)

It took to me some time to understand my mistake. Thank you both for your explanations! 31.154.81.0 (talk) 18:57, 17 July 2016 (UTC)

Lazy people like me will do the partial fraction expansion as follows. The singularity at $y=-1$ gives rise to the term $-{\frac {1}{y+1}}$ the coefficient of -1 here is obtained by taking the limit of $y+1$ times the integrand for y approaching -1, which is trivial to evaluate. The singularity at $y=0$ will produce the remaining part of the partial fraction expansion, we can obtain this effortlessly by using the fact that the integrand decays like $y^{-4}$ for large $y$ . This means that the remaining part of the partial fraction expansion must eliminate the asymptotic terms for large $y$ coming from $-{\frac {1}{y+1}}$ that decay less fast for large $y$ . For large $y$ , we have:

$-{\frac {1}{y+1}}=-{\frac {1}{y}}+{\frac {1}{y^{2}}}-{\frac {1}{y^{3}}}+{\mathcal {O}}\left({\frac {1}{y^{4}}}\right)$

The partial fraction expansion is thus given by:

$-{\frac {1}{y+1}}+{\frac {1}{y}}-{\frac {1}{y^{2}}}+{\frac {1}{y^{3}}}$

Count Iblis (talk) 17:39, 16 July 2016 (UTC)

A simpler solution:

$\int {\frac {1}{e^{2x}+e^{3x}}}dx=-\int {\frac {e^{-2x}}{e^{-x}+1}}d\left(e^{-x}\right)=-\int \left(e^{-x}-1\right)d\left(e^{-x}\right)-\int {\frac {1}{e^{-x}+1}}d\left(e^{-x}\right)=-\left(e^{-2x}/2-e^{-x}\right)-\ln \left|e^{-x}+1\right|+C$

Ruslik_Zero 13:47, 17 July 2016 (UTC)

Thank you all for the detailed solutions! It was really helpful for me! 31.154.81.0 (talk) 18:57, 17 July 2016 (UTC)