How to integrate $\int \frac{1+e^{x}}{1+e^{5x}} dx$?

Question

How to integrate $$\int \frac{1+e^{x}}{1+e^{5x}} dx$$ I am trying to integrate by separating the numerator. So, I am getting $$\int \frac{1}{1+e^{5x}}+\frac{e^{x}}{1+e^{5x}}dx$$ Now I am facing problem to integrate the 1st part $$\int\frac{1}{1+e^{5x}}dx$$ And to solve the 2nd part we have to substitute $e^{x}=u$. So, finally I am getting the 2nd part to be $$\int \frac{1}{1+u^{5}}du$$ But I am facing problem to solve further. Please help me out with this integration.

Answer 1

Given two polynomials $p$ and $q$, you can compute the integral $$ \int\frac{p(x)}{q(x)}\mathrm{d}x $$ by applying the partial fraction decomposition to $\frac{p(x)}{q(x)}$. Assuming all zeros $z_1,\dots,z_n$ of $q$ have multiplicity $1$ and $\forall i=1,\dots,n: p(z_i)\neq 0$, we get $$ \frac{p(x)}{q(x)} = \sum_{i=1}^n \frac{a_i}{x-z_i}. $$ The coefficients $a_i$ can be computed via $$ \frac{p(x)(x-z_i)}{q(x)} = a_i + \sum_{\substack{j=1\\j\neq i}}^n \frac{a_j(x-z_i)}{x-z_j}. $$ Notice we can factor $q(x) = c\prod_{i=1}^n (x-z_i)$ where $c\in\mathbb{C}$ and thus

$$ \frac{p(x)(x-z_i)}{q(x)} = \frac{p(x)}{\prod_{\substack{j=1\\j\neq i}}(x-z_j)} \implies \frac{p(x)(x-z_i)}{q(x)}\Bigg|_{x=z_i}=a_i. $$

Hence given the PFD (partial fraction decomposition) of $\frac{p(x)}{q(x)}$ we can compute $$ \int \frac{p(x)}{q(x)}\mathrm{d}x = \int \sum_{i=1}^n \frac{a_i}{x-z_i}\mathrm{d} x = \sum_{i=1}^n a_i \int\frac{1}{x-z_i}\mathrm{d}x. $$ Using $\int \frac{1}{x-\alpha}\mathrm{d}x = \ln(x-\alpha)$ we get $$ \int \frac{p(x)}{q(x)}\mathrm{d}x = \sum_{i=1}^n a_i \ln(x-z_i). $$

However, there is a problem with this, since $z_i$ may be complex numbers. This requires the use of the complex logarithm. Such is the case for your problem, since $1+u^5 = 0$ has the solutions $$ \zeta_{5,k} = \exp\left(\frac{2\pi\mathrm{i} k}{5}\right) \qquad k=0,\dots,4. $$

This results in $$ \int \frac{1}{1+u^5}\mathrm{d}u = \sum_{k=0}^4 a_k \ln(x - \zeta_{5,k}) $$ This is probably the origin of the nasty solution from WolframAlpha.

Answer 2

To continue, note that the 1st integral is simply $ \int \frac{1}{1+e^{5x}}dx= -\frac15\ln(1+e^{-5x}) $ and integrate the 2nd as follows

\begin{align} &\int \frac5{u^5+1}du\\ =&\ \int \frac1{u+1}+\frac{2(1-u\cos\frac\pi5)}{u^2-2 u\cos\frac\pi5+1}+ \frac{2(1-u\cos\frac{3\pi}5)}{u^2-2u\cos\frac{3\pi}5 +1}\ du\\ =& \ \ln(u+1) +2J(\cos\frac \pi5) +2J(\cos\frac {3\pi}5 ) \end{align} where

\begin{align} J(a)&= \int \frac{1-a u}{u^2-2a u+1}du\\ &=-\frac a2 \ln(u^2-2a u+1)+\sqrt{1-a^2}\tan^{-1}\frac{u-a}{\sqrt{1-a^2}} \end{align}

Answer 3

Since you received the answers for your specific case, consider the more general problem of $$I_{m,n}=\int \frac{1+e^{mx}}{1+e^{nx}}\, dx$$ where $\frac nm$ is not an integer

$$x=\log(t) \qquad \implies \qquad I_{m,n}=\int \frac{t^m+1}{t \left(t^n+1\right)}\,dt$$ which has a solution in terms of the Gaussian hypergeometric function $$I_{m,n}=\frac {t^m} m \,\, _2F_1\left(1,\frac{m}{n};\frac{m+n}{n};-t^n\right)-\frac 1n \log \left(1-t^n\right)$$

Assuming $n>0$ and $m<n$ $$J_{m,n}=\int_0^\infty \frac{1+e^{mx}}{1+e^{nx}}\, dx=\frac{H_{-\frac{m}{2 n}}-H_{-\frac{m+n}{2 n}}+2\log (2)}{2 n}$$

Answer 4

Partial fractions raised by Quanto $$\frac5{u^5+1}=\frac1{u+1}+\frac{2(1-u\cos\frac\pi5)}{u^2-2 u\cos\frac\pi5+1}+ \frac{2(1-u\cos\frac{3\pi}5)}{u^2-2u\cos\frac{3\pi}5 +1}$$ Proof:

Using the identity : $$ u^5+1=(u+1)\left(u^2-2 u \cos \frac{\pi}{5}+1\right)\left(u^2-2 u \cos \frac{3 \pi}{5}+1\right) \tag*{(*)} $$ which can be obtained by solving the quintic equation:$z^5+1=0$.

First of all, let’s transform the equation $(*)$ by the substitution $z=\frac{1}{u}$ as $$ \begin{aligned} & \frac{1}{z^5}+1=\left(\frac{1}{z}+1\right)\left(\frac{1}{z^2}-\frac{2}{z} \cos \frac{\pi}{5}+1\right)\left(\frac{1}{z^2}-\frac{2}{z} \cos \frac{3 \pi}{5}+1\right) \\ \Leftrightarrow \quad & 1+z^5=(1+z)\left(z^2-2 z \cos \frac{\pi}{5}+1\right)\left(z^2-2 z \cos \frac{3 \pi}{5}+1\right) \end{aligned} $$ Applying logarithmic differentiation yields $$ \frac{5 z^4}{1+z^5}=\frac{1}{1+z}+\frac{2\left(z-\cos \frac{\pi}{5}\right)}{z^2-2 z \cos \frac{\pi}{5}+1}+\frac{2\left(z-\cos \frac{3 \pi}{5}\right)}{z^2-2 z \cos \frac{3 \pi}{5}+1} $$

Putting back $z=\frac{1}{u}$ gives $$ \begin{aligned} & \frac{5 \cdot \frac{1}{u^4}}{1+\frac{1}{u^5}}=\frac{1}{1+\frac{1}{u}}+\frac{2\left(\frac{1}{u}-\cos \frac{\pi}{5}\right)}{\frac{1}{u^2}-\frac{2}{u} \cos \frac{\pi}{5}+1}+\frac{2\left(\frac{1}{u}-\cos \frac{3 \pi}{5}\right)}{\frac{1}{u^2}-\frac{2}{u} \cos \frac{3\pi}{5}+1} \\ \Leftrightarrow \quad & \frac{5}{u^5+1}=\frac{1}{u+1}+\frac{2\left(1-u \cos \frac{\pi}{5}\right)}{u^2-2 u \cos \frac{\pi}{5}+1}+\frac{2\left(1-u \cos \frac{3 \pi}{5}\right)}{u^2-2 u\cos \frac{3 \pi}{5}+1} \end{aligned} $$