Ta có:
$\int\limits_1^3\dfrac{1+\ln(x+1)}{x^2}dx$
$=\int\limits_1^3\dfrac{1}{x^2}dx-\int\limits_1^3\ln(x+1)d(\dfrac{1}{x})$
$=-\dfrac{1}{x}\left|\begin{array}{l}3\\1\end{array}\right.-\dfrac{\ln(x+1)}{x}\left|\begin{array}{l}3\\1\end{array}\right.+\int\limits_1^3\dfrac{1}{x}d(\ln(x+1))$
$=\dfrac{2}{3}+\dfrac{\ln2}{3}+\int\limits_1^3\dfrac{1}{x(x+1)}dx$
$=\dfrac{2}{3}+\dfrac{\ln2}{3}+\int\limits_1^3\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)dx$
$=\dfrac{2}{3}+\dfrac{\ln2}{3}+\ln\dfrac{x}{x+1}\left|\begin{array}{l}3\\1\end{array}\right.$
$=\dfrac{2}{3}+\dfrac{\ln2}{3}+\ln\dfrac{3}{2}$