Prove $\int_{0}^{+\infty}\frac{1}{f(x)}dx$ is convergent when $\int_{0}^{+\infty}\frac{e^x}{(e^xf(x))'}dx$ is convergent 
Suppose $f(x)$ is positive monotone increasing function over $[0,\infty)$, and it has derivative. Prove: if $\int_{0}^{+\infty}\frac{e^x}{(e^xf(x))'}dx$ is convergent, then
$\int_{0}^{+\infty}\frac{1}{f(x)}dx$ is convergent.

My work:
$\int_{0}^{+\infty}\frac{e^x}{(e^xf(x))'}=\int_{0}^{\infty}\frac{1}{f(x)+f'(x)}\Bbb{dx}$, but since $f'(x)>0$  I do not know what to do next.
 A: Although I suspect that there might be an easier solution available, this is my best idea so far:

For each $n \in \{0,1,2,\dots\}$, let
$$E_n = \{ x \in [n, n+1] : f'(x) \geq 2f(n+1) \}.$$
Also, let $|E_n|$ denotes the measure of $E_n$.

*

*Assume that $|E_n| > \frac{1}{2}$. Then
$$f(n+1) \geq f(n) + \int_{E_n} f'(x) \, \mathrm{d}x > f(n) + f(n+1),$$
a contradiction.


*The previous step shows that $|E_n| \leq \frac{1}{2}$. Then
\begin{align*}
\int_{n}^{n+1} \frac{\mathrm{d}x}{f(x)+f'(x)}
&\geq \int_{[n,n+1]\setminus E_n} \frac{\mathrm{d}x}{f(x)+f'(x)} \\
&\geq \int_{[n,n+1]\setminus E_n} \frac{\mathrm{d}x}{3f(n+1)} \\
&\geq \frac{1}{6f(n+1)},
\end{align*}
and so,
$$\frac{1}{f(n+1)} \leq 6 \int_{n}^{n+1} \frac{\mathrm{d}x}{f(x)+f'(x)}. $$
Therefore
$$
\int_{0}^{\infty} \frac{\mathrm{d}x}{f(x)}
\leq \sum_{n=0}^{\infty} \frac{1}{f(n)}
\leq \frac{1}{f(0)} + 6 \int_{0}^{\infty} \frac{\mathrm{d}x}{f(x)+f'(x)}.
$$
Now by the assumption, the right-hand side is finite, and therefore $\int_{0}^{\infty} \frac{\mathrm{d}x}{f(x)}$ converges.
