x-3-1-1-3-dx-solution-

Question Number 193426 by 073 last updated on 13/Jun/23

\int \sqrt[3]{x^{3} + 1} dx = ?

solution ?

Answered by witcher3 last updated on 15/Jun/23

x^{3} = t

\int {(x^{3} + 1)}^{\frac{1}{3}} dx = f (x)

\frac{1}{3} \int {(t + 1)}^{\frac{1}{3}} t^{- \frac{2}{3}} dt

{(t + 1)}^{\frac{1}{3}} = 1 + \sum_{k ⩾ 1} \frac{\underset{j = 1}{\prod^{k}} (\frac{4}{3} - j)}{(k)!} t^{k}

\frac{1}{3} \int t^{- \frac{2}{3}} + \sum_{k ⩾ 1} \frac{{(- 1)}^{k} Γ (k - \frac{1}{3})}{Γ (k + 1) Γ (- \frac{1}{3})} t^{k - \frac{2}{3}} dt

= t^{\frac{1}{3}} + \frac{1}{3} \sum_{k ⩾ 1} \frac{Γ (k - \frac{1}{3}) {(- 1)}^{k}}{Γ (- \frac{1}{3}) (k + \frac{1}{3})} \frac{t^{k + \frac{1}{3}}}{k!} + c

t^{\frac{1}{3}} (1 + \frac{1}{3} \sum_{k ⩾ 1} \frac{Γ (k - \frac{1}{3})}{Γ (- \frac{1}{3})} . \frac{Γ (k + \frac{1}{3})}{Γ (k + \frac{4}{3})} . \frac{{(- t)}^{k}}{k!}) + c

\frac{1}{3} = \frac{Γ (\frac{4}{3})}{Γ (\frac{1}{3})}

\Leftrightarrow t^{\frac{1}{3}} (1 + \sum_{k ⩾ 1} \frac{Γ (k - \frac{1}{3})}{Γ (- \frac{1}{3})} . \frac{\frac{Γ (k + \frac{1}{3})}{Γ (\frac{1}{3})}}{\frac{Γ (k + \frac{4}{3})}{Γ (\frac{4}{3})}} . \frac{{(- t)}^{k}}{k!}) + c

{(a)}_{k} = \frac{Γ (a + k)}{Γ (a)}

t^{\frac{1}{3}} (1 + \sum_{k ⩾ 1} \frac{{(- \frac{1}{3})}_{k} {(\frac{1}{3})}_{k}}{{(\frac{4}{3})}_{k}} . \frac{{(- t)}^{k}}{k!}) + c

t^{\frac{1}{3}}_{2} F_{1} (- \frac{1}{3}, \frac{1}{3}; \frac{4}{3}, - t) + c

f (x) = x_{2} F_{1} (- \frac{1}{3}, \frac{1}{3}; \frac{4}{3}, - x^{3}) + c, c \in R

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