#### Answer

$y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+1)$

#### Work Step by Step

Here, we have $y'+\dfrac{2y}{x+1}=\dfrac{x}{x+1}$
The integrating factor is: $I=e^{\int (\dfrac{2}{x+1}) dx}=(x+1)^2$
Now, $(x+1)^2[y'+\dfrac{2}{x+1} y]=(x+1)^2(\dfrac{x}{x+1})$
$\int [(x+1)^2y]' =\int x(x+1) dx$
This implies that
$(x+1)^2 (y)=\int (x^2+x) dx$
$\implies y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+c)$
Further, apply the initial conditions.
$y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+c) \implies y=(x+1)^{-2}(\dfrac{x^{3}}{3}+\dfrac{x^2}{2}+1)$