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Question Number 130725 by mathmax by abdo last updated on 28/Jan/21

$$\mathrm{find}{\int }_0^1\frac{{\mathrm{x}}^2}{\sqrt{{\mathrm{x}}^4+1}}\mathrm{dx}$$

Answered by Dwaipayan Shikari last updated on 28/Jan/21

$$\frac{1}{\sqrt{1+x}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{(-1)}^n{x}^n$$$$\frac{x^2}{\sqrt{1+x^4}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!}{(-1)}^n{x}^{4n+2}$$$${\int }_0^1\frac{x^2}{\sqrt{1+x^4}}=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n}{n!(4n+3)}{(-1)}^n=\frac{1}{3}\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{\left(\frac{1}{2}\right)}_n{\left(\frac{3}{4}\right)}_n}{{\left(\frac{7}{4}\right)}_{n}n!}{(-1)}^n$$$$As,\frac{1}{4n+3}=\frac{1}{4}\left(\frac{1}{n+\frac{3}{4}}\right)=\frac{1}{4}\left(\frac{\mathrm{\Gamma }(n+\frac{3}{4})}{\mathrm{\Gamma }(n+\frac{7}{4})}\right)=\frac{4}{3}.\frac{1}{4}.\frac{{\left(\frac{3}{4}\right)}_n}{{\left(\frac{7}{4}\right)}_n}$$$$=\frac{1}{3}{}_2{F}_1(\frac{1}{2},\frac{3}{4};\frac{7}{4};-1)$$

Commented by mathmax by abdo last updated on 28/Jan/21

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

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