Question

Encontre a integral indefinida ∫ x 2 x + 1 d x

Answer 1

$\boxed{I=\int{\dfrac{x^2}{x+1}dx}}$

$u=x+1\ \therefore\ \dfrac{du}{dx}=1\ \therefore\ du=dx$

$x=u-1\ \therefore\ x^2=(u-1)^2$

$I=\int{\dfrac{(u-1)^2}{u}du}=\int{\dfrac{u^2-2u+1}{u}du}\ \therefore$

$I=\int{\bigg(u-2+\dfrac{1}{u}\bigg)du}=\int{udu}-2\int{du}+\int{\dfrac{1}{u}du}\ \therefore$

$I=\dfrac{u^2}{2}-2u+\ln{|u|}+C\ \therefore$

$\boxed{I=\dfrac{(x+1)^2}{2}-2(x+1)+\ln{|x+1|}+C}$

Letra D.