Answer
$\dfrac{1}{x}+\ln (\dfrac{x-1}{x})^2+c$
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: $\int \dfrac{(x+1)dx}{x^2(x-1)}=\int [\dfrac{-2}{x}-\dfrac{1}{x^2}+\dfrac{2}{x-1} ]dx$
This implies that $-2\ln |x|+\dfrac{1}{x}+2\ln |x-1|+c=\dfrac{1}{x}+2\ln |\dfrac{x-1}{x}|+c$
Hence, $\dfrac{1}{x}+2\ln |\dfrac{x-1}{x}|+c=\dfrac{1}{x}+\ln (\dfrac{x-1}{x})^2+c$
