Integration of an improper integral

Question

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}

Answer 1

Invoking the definition of the improper integral;

\begin{align*} \int_1^\infty \left ( \frac{1}{x} - \frac{1}{s+x} \right ) \, \mathrm{d}x &= \lim_{y \rightarrow +\infty} \int_{1}^{y} \left ( \frac{1}{x} - \frac{1}{s+x} \right ) \, \mathrm{d}x \\ &=\lim_{y\rightarrow +\infty} \left ( \ln y - \ln \left ( \frac{s+y}{s} \right ) + \ln \left ( \frac{1}{s}+ 1 \right ) \right ) \end{align*}

Can you take it from here?

Final answer is $\ln (1+s)$.

Answer 2

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

$=\ln (\frac M {M+s})+\ln(1+s)$.

$Let$ $M \to \infty$.