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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:

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

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:

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

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)