Integration of an improper integral I have an integral that has two individually divergent parts. Wolfram says that the answer is \begin{equation}\frac{\ln(s+1)}{s},\end{equation} but I cannot figure out how it's done. The integral is
\begin{equation}
\int_{1}^{\infty}\left({\frac{1}{x}-\frac{1}{x+s}}\right)dx
\end{equation}

$\int_1^{M} (\frac 1 x -\frac 1 {x+s})dx =\ln M-[\ln (M+s)-\ln (1+s)]$
$=\ln (\frac M {M+s})+\ln(1+s)$.
$Let$ $M \to \infty$.
