How to prove that this improper integral does not converge? Prove that
$$\int_1^\infty\frac{e^x}{x (e^x+1)}dx$$
does not converge.
How can I do that? I thought about turning it into the form of $\int_b^\infty\frac{dx}{x^a}$, but I find no easy way to get rid of the $e^x$.
 A: Note that the integrand is positive. Since
$$
\lim_{x\to\infty} \frac{e^x}{e^x+1} = 1,
$$
it follows that there is an $X$ such that  $\dfrac{e^x}{e^x+1} \ge \dfrac12$ for $x \ge X$. Hence
\begin{align*}
\int_1^\infty \frac{e^x}{x(e^x+1)}\,dx &= \int_1^X \frac{e^x}{x(e^x+1)}\,dx + \int_X^\infty \frac{e^x}{x(e^x+1)}\,dx \\
&\ge \int_1^X \frac{e^x}{x(e^x+1)}\,dx + \frac12 \int_X^\infty \frac{1}{x}\,dx.
\end{align*}
The second of these integrals diverges (why?), so the original integral is also divergent.
A: If we divide the top and bottom by $e^x$, we have
$$\int_{1}^{\infty}\frac{1}{x+x/e^x}$$
For large values of $x$, $x/e^x < x$, so $x+x/e^x < 2x$ and therefore $1/(x+x/e^x) > 1/2x$. Then the tail of $1/2x$ lies under the curve of $1/(x+x/e^x)$. Then since
$$\int_{1}^{\infty}\frac{1}{2x}$$
diverges, we know that the first integral diverges as well.
A: Collect $e^x$ and have
$$\int\frac{e^x}{e^x(x + e^{-x})} = \int \frac{dx}{x + xe^{-x}}$$
Now substitute $e^{-x} = y$ so that $dy = -y dx$ and extrema changes from $1/e$ to $0$:
$$\int_0^{1/e} \frac{dy}{y(- \log(y))(1+y)}$$
which has a pole along the path of integration, so then the integral does not converge.
A: \begin{align}
&\int_1^{\infty}{e^x\over x(e^x+1)}dx,\quad e^x+1\mapsto y\\
&=\int_{e+1}^{\infty}{1\over y\ln (y-1) }dy>\int_{e+1}^{\infty}{1\over y\ln y }dy=\left[\ln (\ln y)\right]_{e+1}^\infty=\infty
\end{align}
A: Since $~\dfrac{dx}x=d~\big(\ln x\big),~$ we'll just let $x=e^t.~$ This yields $\displaystyle\int_0^\infty\frac{e^{e^t}}{e^{e^t}+1}~dt.~$ Simplifying 
both sides by $e^{e^t}$ leads to $\displaystyle\int_0^\infty\frac1{1+e^{-e^t}}~dt,~$ which, as far as I'm able to see, diverges 
as shamelessly as $\displaystyle\int_0^\infty dt,~$ since the rate at which the function $e^{-e^t}$ decreases towards 
$0$ is simply mind-blowing.
A: For $x\geq1$, you have $e^x>1$, so $2e^x>e^x+1$, or 
$$
\frac {e^x}{e^x+1}>\frac12.
$$
Then 
$$
\int_1^\infty \frac {e^x}{x (e^x+1)}\,dx\geq\int_1^\infty\frac1x\,dx=\infty .
$$
