Question

Usando o método de integração por partes: ∫ u .dv = u .v- ∫ v.du, e lembrando que (ln x)’ =1/x,determine I= ∫ ln x.5x^4 dx

Answer 1

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Solução!

$\displaystyle \int (lnx.5x^{4})dx\\\\\ u=lnx~~~~du= (\frac{1}{x})dx \\\\\ dv=5 x^{4} ~~~~~v= \frac{5 x^{5} }{5} ~~~~v= x^{5}$

$\displaystyle \int (lnx.5x^{4})=u.v-\displaystyle \int v.du\\\\\\\\\\ \displaystyle \int (lnx.5x^{4})=lnx. x^{5} -\displaystyle \int (x^{5} . \frac{1}{x}) dx\\\\\\\\\ \displaystyle \int (lnx.5x^{4})=lnx. x^{5} -\displaystyle \int ( \frac{ x^{5} }{x}) dx\\\\\\\\\ \displaystyle \int (lnx.5x^{4})=lnx. x^{5} -\displaystyle \int ( x^{4} ) dx\\\\\\\\\ \displaystyle \int (lnx.5x^{4})=lnx. x^{5} - \frac{ x^{5} }{5}+c$


$\boxed{Resposta:\displaystyle \int (lnx.5x^{4})dx=lnx. x^{5} - \frac{ x^{5} }{5}+c}$

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