Answer
$$ - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| {\frac{x}{{x + 1}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = \ln \left( {x + 1} \right),{\text{ }}du = \frac{1}{{x + 1}} \cr
& dv = \frac{1}{{{x^2}}},{\text{ }}v = - \frac{1}{x} \cr
& {\text{Using integration by parts formula}} \cr
& \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \int {\frac{1}{{x\left( {x + 1} \right)}}} dx \cr
& {\text{Decomposing }}\frac{1}{{x\left( {x + 1} \right)}}{\text{ into partial fractions }} \cr
& \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \int {\left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| x \right| - \ln \left| {x + 1} \right| + C \cr
& \int {\frac{{\ln \left( {x + 1} \right)}}{{{x^2}}}} dx = - \frac{1}{x}\ln \left( {x + 1} \right) + \ln \left| {\frac{x}{{x + 1}}} \right| + C \cr} $$