Ta có:
$\int\limits_1^3\dfrac{3+\ln x}{(x+1)^2}dx$
$=\int\limits_1^3\dfrac{3}{(x+1)^2}dx-\int\limits_1^3\ln xd(\dfrac{1}{x+1})$
$=-\dfrac{3}{x+1}\left|\begin{array}{l}3\\1\end{array}\right.-\dfrac{\ln x}{x+1}\left|\begin{array}{l}3\\1\end{array}\right.+\int\limits_1^3\dfrac{1}{x+1}d(\ln x)$
$=\dfrac{3}{4}-\dfrac{\ln 3}{4}+\int\limits_1^3\dfrac{1}{x(x+1)}dx$
$=\dfrac{3}{4}-\dfrac{\ln 3}{4}+\int\limits_1^3\left(\dfrac{1}{x}-\dfrac{1}{x+1}\right)dx$
$=\dfrac{3}{4}-\dfrac{\ln 3}{4}+\ln\dfrac{x}{x+1}\left|\begin{array}{l}3\\1\end{array}\right.$
$=\dfrac{3}{4}-\dfrac{\ln 3}{4}+\ln\dfrac{3}{2}$