Question

Calcule a integral (1+p)/∛p dp

Answer 1

$\mathrm{\int\dfrac{1+p}{\sqrt[3]{p}}\ dp=\int\dfrac{1}{p^{\frac{1}{3}}}+\dfrac{p}{p^{\frac{1}{3}}}\ dp=}\\\\\\ \mathrm{=\int p^{-\frac{1}{3}}\ dp+\int p^{\frac{2}{3}}\ dp=\dfrac{p^{-\frac{1}{3}+1}}{-\frac{1}{3}+1}+\dfrac{p^{\frac{2}{3}+1}}{\frac{2}{3}+1}=}\\\\\\ \mathrm{=\dfrac{p^{\frac{2}{3}}}{\frac{2}{3}}+\dfrac{p^{\frac{5}{3}}}{\frac{5}{3}}=\dfrac{3p^{\frac{2}{3}}}{2}+\dfrac{3p^{\frac{5}{3}}}{5}=\dfrac{15p^{\frac{2}{3}}+6p^{\frac{5}{3}}}{10}=}$

$\mathrm{=\dfrac{15p^{\frac{2}{3}}+6p^{\frac{2}{3}}p}{10}=\mathbf{\dfrac{3p^{\frac{2}{3}}(2p+5)}{10}}}$

Answer 2

$\displaystyle\mathsf{\int\dfrac{1+p}{\sqrt[3]{p}} dp = \int {p}^{ - \frac{1}{3}}(1 + p)dp} \\ \displaystyle\mathsf{\int ({p}^{ - \frac{1}{3} } + {p}^{ \frac{2}{3}} )dp =\dfrac{2}{3} {p}^{ \frac{3}{2}} + \frac{3}{5} {p}^{ \frac{5}{3} } + k}$