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

= 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)