Question

alguem me ajude com esta integral por partes

Answer 1

Resolução da questão, vejamos:

Resolver a integral:

$\mathsf{\displaystyle\int~\dfrac{e^{2x}}{e^{x}+4}}}~\mathsf{dx}}}$

Suponhamos que:

$\mathsf{u=e^{x}}}}~=>~\mathsf{du=e^{x}~dx}}}$

Observemos:

$\mathsf{\displaystyle\int~\dfrac{u}{u+4}}~\mathsf{du}}~=>~\mathsf{1-\dfrac{4}{u+4}}}$

Efetuemos a integração por partes:

$\mathsf{\displaystyle\int~1~du=u}}}}\\\\\\\\ \mathsf{\displaystyle\int~\dfrac{4}{u+4}}~\mathsf{du}}}=\mathsf{-4~\displaystyle\int~\dfrac{1}{u+4}}~\mathsf{du}}}}$

Suponhamos que:

u = u + 4 => du = du

Vejamos:

$\mathsf{\displaystyle\int~\dfrac{1}{u}}~\mathsf{du}}=\mathsf{\displaystyle\int~log(u)~du}}}\\\\\\\\ \mathsf{log~(u+4)}}}\\\\\\\ \mathsf{u-4log(u+4)}}\\\\\\\ \mathsf{e^{x}-4log(e^{x}+4)}}$

Assim sendo, concluímos que:

$\Large\boxed{\boxed{\boxed{\mathbf{\displaystyle\int~\dfrac{e^{2x}}{e^{x}+4}}~\mathbf{dx}=\mathbf{e^{x}-4log(e^{x}+4)+C}}}}}}}}}}}}}}}}~~\checkmark}}}$

Espero que te ajude (^.^)