Solve the integration by parts

Question

Solve this integration by parts

∫ln(x+1)/x dx

Answer 1

Let us make the change $x = -u$. Then, the integral becomes $$ \begin{align*}\int \frac{\ln(x+1)}{x} dx &= \int \frac{\ln(1 - u)}{-u} d(-u) \\ &= - \int \frac{\ln(1 - u)}{-u} du\end{align*}$$

The value $f(z) = \int_0^z \frac{\ln(1 - u)}{-u} du $ is a standard complex function called Dilogarithm, which is denoted as $Li_2(z)$ (check this link for more information). To conclude your exercise, $$ \begin{align*}- \int \frac{\ln(1 - u)}{-u} du &= - Li_2(u) + C \\ &= - Li_2(-x) + C\end{align*}$$ Hence, $$ \begin{align*}\int \frac{\ln(x+1)}{x} dx = - Li_2(-x) + C\end{align*}$$