Answer
$$ -\frac{\ln (x+1)}{x}+ \ln (x)-\ln(x+1)+c$$
Work Step by Step
Given
$$\int \frac{\ln (x+1)}{x^2} d x$$
Use integration by parts
\begin{aligned}
u&= \ln (x+1) \ \ \ \ \ \ dv= x^{-2}dx\\
du&= \frac{dx}{x+1}\ \ \ \ \ \ \ \ \ v= \frac{-1}{x}
\end{aligned}
Then
\begin{aligned}
\int \frac{\ln (x+1)} {x^2} d x&= -\frac{\ln (x+1)}{x}+\int \frac{dx}{x(x+1)}
\end{aligned}
Use partial fractions
\begin{aligned}
\frac{1}{x\left(x+1\right)}&=\frac{a_0}{x}+\frac{a_1}{x+1}\\
1&= a_0\left(x+1\right)+a_1x
\end{aligned}
at $x=0\to\ \ a_0 = 1,\ \ x=-1\to\ \ a_0 = -1$
Then
\begin{aligned}
\int \frac{dx}{x(x+1)}&= \int \frac{1}{x}dx-\int \frac{1}{x+1}dx\\
&= \ln (x)-\ln(x+1)+c
\end{aligned}
Hence
\begin{aligned}
\int \frac{\ln (x+1)} {x^2} d x&= -\frac{\ln (x+1)}{x}+ \ln (x)-\ln(x+1)+c
\end{aligned}
