Answer
$\ln |\dfrac{x}{x+1}|+\dfrac{1}{x+1} +c$
Work Step by Step
Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$
Write the integral into partial fractions as follows: $\int \dfrac{dx}{x(x+1)^2}=\int [\dfrac{1}{x}-\dfrac{1}{x+1}-\dfrac{1}{(x+1)^2} ]dx$
Hence, $\ln |x|-\ln |x+1|+\dfrac{1}{x+1}+c=\ln |\dfrac{x}{x+1}|+\dfrac{1}{x+1} +c$
