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$I= \int_{1}^{3}\frac{3+\ln x}{(1+x)^2}dx=\int_{1}^{3}\frac{3}{(1+x)^2}dx+\int_{1}^{3}\frac{\ln x}{(1+x)^2}dx=\left[ {-\frac{3}{1+x}} \right]_{1}^{3}+\int_{1}^{3}\frac{\ln x}{(1+x)^2}dx$
$=\dfrac{3}{4}+\int_{1}^{3}\frac{\ln x}{(1+x)^2}dx$
Đặt $\begin{cases}u=\ln x \\ dv=\frac{1}{(1+x)^2}dx \end{cases}\Leftrightarrow \begin{cases}du=\dfrac{1}{x}dx \\ v=-\frac{1}{1+x}dx \end{cases}$
Suy ra
$I=\dfrac{3}{4}+\left[ {-\frac{\ln x}{1+x}} \right]_{1}^{3}+\int_{1}^{3}\frac{1}{x(1+x)}dx$
$I=\dfrac{3}{4}+\left[ {-\frac{\ln x}{1+x}} \right]_{1}^{3}+\int_{1}^{3}\left ( \frac{1}{x}-\frac{1}{1+x} \right )dx$
$I=\dfrac{3}{4}+\left[ {-\frac{\ln x}{1+x}+\ln x -\ln (x+1)} \right]_{1}^{3}$
$I=\boxed{\dfrac{3}{4}+\dfrac{1}{4}\ln\dfrac{27}{16}} $
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