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Question Number 82891 by jagoll last updated on 25/Feb/20

$$\int \frac{1}{\sqrt{1-{\mathrm{x}}^4}}\mathrm{dx}=?$$

Answered by mind is power last updated on 25/Feb/20

$$\frac{1}{.\sqrt{1-t^2}}=\sum _{n⩾0}\frac{\left(2n\right)!t^{2n}}{2^{2n}{(n!)}^2}$$$$\frac{1}{\sqrt{1-t^4}}=\sum _{n⩾0}\frac{\left(2n\right)!t^{4n}}{2^{2n}{(n!)}^2}$$$$\int \frac{dt}{\sqrt{1-t^4}}=\sum _{n⩾0}\frac{\left(2n\right)!}{2^{2n}{(n!)}^2}\int t^{4n}dt$$$$=\sum _{n⩾0}\frac{\left(2n\right)!x^{4n+1}}{2^{2n}{(n!)}^2(4n+1)}=x\sum _{n⩾0}\frac{\left(2n\right)!x^{4n}}{2^{2n}{(n!)}^2(4n+1)}$$$$x\sum _{n⩾0}\frac{2^n.n!.(2n+1)!!}{2^{2n}{(n!)}^2(4n+1)}{x}^{4n}$$$$=x\sum _{n⩾0}\frac{(2n+1)!!}{2^n(4n+1)}.\frac{{\left(x^4\right)}^n}{n!}+c$$$$(2n+1)!!=2^n\underset{k=0}{\prod ^{n-1}}(k+\frac{3}{2})=2^n{\left(\frac{3}{2}\right)}_n$$$$=x\sum _{n⩾0}\frac{{\left(\frac{3}{2}\right)}_n.2^n\left(\frac{1}{4}\right)}{2^n.(n+\frac{1}{4})}.x^{4n}+c$$$$=x\sum _{n⩾0}\frac{{\left(\frac{3}{2}\right)}_n.{\left(\frac{1}{4}\right)}_n}{{\left(\frac{5}{4}\right)}_n}.x^{4n}+c$$$$=x{}_2{F}_1(\frac{3}{2},\frac{1}{4};\frac{5}{5};x^4)+c$$$$$$$$$$

Commented by M±th+et£s last updated on 25/Feb/20

$$greatsir$$

Commented by jagoll last updated on 26/Feb/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}$$

Answered by M±th+et£s last updated on 25/Feb/20

$$I=\int \frac{dx}{\sqrt{1+x^2}\sqrt{1-x^2}}$$$$=\int \frac{\frac{1}{\sqrt{1-x^2}}}{\sqrt{1-(-1)(sin{\left({sin}^{-1}\left(x\right)\right)}^2}}dx$$$$=F({sin}^{-1}x\mid -1)+c$$

Commented by M±th+et£s last updated on 25/Feb/20

$$notice(u\mid a)is1stkindelipticalintrgral$$

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