$\int \frac{x^2+3 x-1}{(x+1)^2} d x$ is equal to

$x+\frac{3}{x+1}+\log _e|x+1|+C$

$I = \int \frac{x^2+3 x-1}{(x+1)^2} d x$

so  $\frac{x^2+3 x-1}{(x+1)^2}=\frac{x^2+3 x-1}{x^2+2 x+1}$

$\Rightarrow \frac{x^2+2 x+1+x-2}{x^2+2 x+1}=1+\frac{x-2}{x^2+2 x+1}$

$\Rightarrow 1+\frac{(x+1)-3}{(x+1)^2}=1+\frac{1}{(x+1)}-\frac{3}{(x+1)^2}$

$\Rightarrow I=\int 1+\frac{1}{(x+1)} \frac{-3}{(x+1)^2} d x$

$= x+\log |x+1|+\frac{3}{(x+1)}+C$

Option : B