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Question Number 161830 by EnterUsername last updated on 22/Dec/21

$$\int \frac{dx}{\sqrt{1-x^{14}}}$$

Answered by Ar Brandon last updated on 22/Dec/21

$$\arcsin \left(x\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{2n+1}{\left(\frac{1}{2}\right)}_n}{n!(2n+1)}\Rightarrow \arcsin \left(x^7\right)=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{14n+7}{\left(\frac{1}{2}\right)}_n}{n!(2n+1)}$$$$\Rightarrow \frac{7x^6}{\sqrt{1-x^{14}}}=7\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{x^{14n+6}{\left(\frac{1}{2}\right)}_n}{n!}\Rightarrow \frac{1}{\sqrt{1-x^{14}}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{x}^{14n}$$$$I=\int \frac{dx}{\sqrt{1-x^{14}}}=\int \underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{x}^{14n}dx=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(14n+1)}{x}^{14n+1}$$$$=\frac{x}{14}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(n+\frac{1}{14})}{x}^{14n}=\frac{x}{14}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n\mathrm{\Gamma }(n+\frac{1}{14})}{n!\mathrm{\Gamma }(n+\frac{15}{14})}{x}^{14n}$$$$=\frac{x}{14}\cdot \frac{\mathrm{\Gamma }\left(\frac{1}{14}\right)}{\mathrm{\Gamma }\left(\frac{15}{14}\right)}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n{\left(\frac{1}{14}\right)}_n}{n!{\left(\frac{15}{14}\right)}_n}{x}^{14n}=x\underset{2}{}{F}_1(\frac{1}{2},\frac{1}{14};\frac{15}{14};x^{14})$$

Commented by Lordose last updated on 22/Dec/21

$$\mathrm{fast}$$

Commented by Ar Brandon last updated on 22/Dec/21

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