Calcule a integral

por José Fernandes de Brito Qui 22 Mar 2018, 09:32

Calcule a integral.

∫(e^(t )+ 2)^(1/3) e^2t dt

por evandronunes Qui 22 Mar 2018, 12:02

Seja \int (e^t+ 2)^{1/3} .e^{2t} dt .

Faça u = e^t+ 2, logo du = e^t dt ou dt = e^{-t} du=(e^{t})^{-1} du=(u-2)^{-1} du.
Temos que e^{2t} =(u-2)^2.

Substituindo,

\int (u)^{1/3} .(u-2)^2 .(u-2)^{-1} du =

=\int (u)^{1/3} .(u-2) du =

=\int (u^{4/3}-2u^{1/3}) du =

=\int u^{4/3} du - \int 2u^{1/3} du =

=\frac{3u^{7/3}}{7} - \frac{3u^{4/3}}{2}

Portanto,

\int (e^t+ 2)^{1/3} .e^{2t} dt=\frac{3(e^t+ 2)^{7/3}}{7} - \frac{3(e^t+ 2)^{4/3}}{2} + C

Calcule a integral

Calcule a integral

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