## Thomas' Calculus 13th Edition

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