[tex]\boxed{I=\int{\dfrac{x^2}{x+1}dx}}[/tex]
[tex]u=x+1\ \therefore\ \dfrac{du}{dx}=1\ \therefore\ du=dx[/tex]
[tex]x=u-1\ \therefore\ x^2=(u-1)^2[/tex]
[tex]I=\int{\dfrac{(u-1)^2}{u}du}=\int{\dfrac{u^2-2u+1}{u}du}\ \therefore[/tex]
[tex]I=\int{\bigg(u-2+\dfrac{1}{u}\bigg)du}=\int{udu}-2\int{du}+\int{\dfrac{1}{u}du}\ \therefore[/tex]
[tex]I=\dfrac{u^2}{2}-2u+\ln{|u|}+C\ \therefore[/tex]
[tex]\boxed{I=\dfrac{(x+1)^2}{2}-2(x+1)+\ln{|x+1|}+C}[/tex]
Letra D.
