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Question Number 180379 by cortano1 last updated on 11/Nov/22

$$\int x^2\sqrt[6]{x^6+5}dx=?$$

Answered by Ar Brandon last updated on 11/Nov/22

$$I=\int x^2\sqrt[6]{x^6+5}dx=\sqrt[6]{5}\int x^2\sqrt[6]{1+{\left(\frac{x}{\sqrt[6]{5}}\right)}^6}dx$$$$\sqrt[6]{1+x}=1+\frac{1}{6}x+\frac{\left(\frac{1}{6}\right)(-\frac{5}{6})x^2}{2!}+\frac{\left(\frac{1}{6}\right)(-\frac{5}{6})(-\frac{11}{6})x^3}{3!}+\cdot \cdot \cdot$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-x)}^n}{n!}\cdot \frac{\mathrm{\Gamma }(n-\frac{1}{6})}{\mathrm{\Gamma }(-\frac{1}{6})}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-x)}^n}{n!}{(-\frac{1}{6})}_n$$$$I=\sqrt[6]{5}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!}{(-\frac{1}{6})}_n\int x^2{\left(\frac{x^6}{5}\right)}^{n}dx=\sqrt[6]{5}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n!(6n+3)}{(-\frac{1}{6})}_n\frac{1}{5^n}{x}^{6n+3}+C$$$$=\sqrt[6]{5}\frac{x^3}{6}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n\mathrm{\Gamma }(n+\frac{1}{2})}{n!(n+\frac{1}{2})\mathrm{\Gamma }(n+\frac{1}{2})}{(-\frac{1}{6})}_n\frac{1}{5^n}{x}^{6n}+C=\sqrt[6]{5}\frac{x^3}{6}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\mathrm{\Gamma }(n+\frac{1}{2})}{n!\mathrm{\Gamma }(n+\frac{3}{2})}{(-\frac{1}{6})}_n{(-\frac{x^6}{5})}^n+C$$$$=\sqrt[6]{5}\frac{x^3}{6}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n\mathrm{\Gamma }\left(\frac{1}{2}\right)}{n!{\left(\frac{3}{2}\right)}_n\mathrm{\Gamma }\left(\frac{3}{2}\right)}{(-\frac{1}{6})}_n{(-\frac{x^6}{5})}^n+C=\sqrt[6]{5}\frac{x^3}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n{(-\frac{1}{6})}_n}{n!{\left(\frac{3}{2}\right)}_n}{(-\frac{x^6}{5})}^n+C$$$$=\frac{1}{3}\sqrt[6]{5}{x}^3\underset{2}{}{F}_1(-\frac{1}{6},\frac{1}{2};\frac{3}{2}\mid -\frac{x^6}{5})+C$$

Commented by cortano1 last updated on 12/Nov/22

$$\mathrm{waw}$$

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