Integral $\int_{0}^{+\infty}\frac{1-e^{-x}-\log(1+x)}{x^2}dx$. 
$$\int_{0}^{+\infty}\frac{1-e^{-x}-\log(1+x)}{x^2}dx$$

Though of Using Maclaurin for log but I don't think it will get me anywhere and nor does the 1/x substitution work. Any hint?
 A: As with Jack D'Aurizio's solution, we begin by integrating by parts with $u=1-e^{-x}-\log(1+x)$ and $v=\frac1x$ to obtain
$$\begin{align}
\int_0^\infty \frac{1-e^{-x}-\log(1+x)}{x^2}\,dx&=\int_0^\infty \frac{e^{-x}-\frac1{1+x}}{x}\,dx\\\\
&=\lim_{\epsilon\to0^+}\left(\int_\epsilon^\infty \frac{e^{-x}}{x}\,dx-\int_\epsilon^\infty \frac{1}{x(1+x)}\,dx\right)\tag1
\end{align}$$
We integrate by parts the first integral on the right-hand side of $(1)$ to reveal
$$\begin{align}
\int_0^\infty \frac{1-e^{-x}-\log(1+x)}{x^2}\,dx&=\lim_{\epsilon\to0^+}\left(e^{-\epsilon}\log(\epsilon)+\int_\epsilon^\infty e^{-x}\log(x)\,dx-\log(\epsilon)-\log(1+\epsilon)\right)\\\\
&=\int_0^\infty e^{-x}\log(x)\,dx\tag2\\\\
&=-\gamma
\end{align}$$
And we are done!


NOTE:  I showed in the Note at the end of THIS ANSWER, the equivalence of the integral representation of 
$$-\gamma=\int_0^\infty e^{-x}\log(x)\,dx$$
in $(2)$ and the limit representation 
$$-\gamma=\lim_{N\to \infty}\left(-\log(N)+\sum_{n=1}^N\frac{1}{n}\right)$$

A: Hint  :  $\frac {\frac x{1+x} - ln(1+x)}{x^2} = \frac d{dx} \{ \frac {ln(1+x)}x \}$
Then write it as $\int_{0}^ \infty \frac {1-e^{-x} - {\frac x{1+x}}}{x^2} + $ $\int_{0}^ \infty \ d \left (\frac {ln(1+x)}x \right ) $
A: The issue in using the Maclaurin series for $\log(1+x)$ is that its radius of convergence is one, while the integration range extends past $1$. On the other hand, by integration by parts, the given integral equals
$$ \int_{0}^{+\infty}\left(e^{-x}-\frac{1}{x+1}\right)\frac{dx}{x} $$
which is a well-known integral representation for the opposite of the Euler-Mascheroni constant, $\color{blue}{-\gamma}$.
Here it is a proof through the usual definition of $\gamma$. A useful lemma is $\int_{0}^{+\infty}\frac{e^{-ax}-e^{-bx}}{x}\,dx = \log\frac{b}{a}$ for any $a,b>0$, which is Frullani's integral. We have
$$ \gamma=\lim_{n\to +\infty} H_n-\log n = \sum_{n\geq 1}\left[\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right]$$
hence the RHS can be represented as
$$ \int_{0}^{+\infty}\sum_{n\geq 1} e^{-nx}-\frac{e^{-nx}-e^{-(n+1)x}}{x}\,dx =\int_{0}^{+\infty}\left(\frac{1}{x e^x}-\frac{1}{e^x-1}\right)\,dx$$
and it is enough to show that
$$ \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x(x+1)}\right)\,dx=0, $$
which follows from $\int\frac{dx}{x(x+1)}=C+\log(x)-\log(1+x)$ and $\int\frac{dx}{e^x-1}=C+\log(e^x-1)-x.$
