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Question Number 122157 by mathmax by abdo last updated on 14/Nov/20

$$\mathrm{find}{\int }_{-1}^1\sqrt{1+{\mathrm{x}}^4}\mathrm{dx}$$

Commented by peter frank last updated on 14/Nov/20

$$\mathrm{Qn}121384$$

Answered by mindispower last updated on 15/Nov/20

$$=2{\int }_0^1\sqrt{1+x^4}dx$$$$x^4=u\Rightarrow dx=u^{-\frac{3}{4}}\frac{du}{4}$$$$=\frac{1}{2}{\int }_0^1\sqrt{1+u}.u^{-\frac{3}{4}}du=I$$$$\beta {(b,c-b)}_2{F}_1(a,b,c;z)={\int }_0^1{x}^{b-1}{(1-x)}^{c-b-1}{(1-zx)}^{-a}dx$$$$\beta bettafunction{,}_2{F}_1(a,b,c;z)Hypergeometric$$$$function$$$$I\iff \frac{1}{2}{\int }_0^1{(1+u)}^{\frac{1}{2}}{u}^{-\frac{3}{4}}{(1-u)}^{0}du$$$$\Rightarrow b-1=-\frac{3}{4},c-b-1=0,-a=\frac{1}{2},-z=1$$$$\Rightarrow b=\frac{1}{4},c=\frac{5}{4},a=-\frac{1}{2},z=-1$$$$I=\frac{1}{2}\beta (\frac{1}{4},1){}_2{F}_1(-\frac{1}{2},\frac{1}{4},\frac{5}{4};-1)$$$$\frac{\beta (\frac{1}{4},1)}{2}=\frac{1}{2}\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }(1+\frac{1}{4})}=2$$$$weget{\int }_{-1}^1\sqrt{1+x^4}dx=2{\int }_0^1\sqrt{1+x^4}dx=2{.}_2{F}_1(-\frac{1}{2},\frac{1}{4},\frac{5}{4};-1)$$

Answered by Dwaipayan Shikari last updated on 15/Nov/20

$${\int }_{-1}^1\sqrt{1+x^4}dx$$$$=2{\int }_0^1\sqrt{1-t}dt{x}^4=-t\Rightarrow 4x^3=-\frac{dt}{dx}$$$$=\frac{1}{2}{\int }_0^1{x}^{-3}\sqrt{1-t}dt$$$$=\frac{1}{2}{\int }_0^1{(-1)}^{-\frac{3}{4}}{t}^{-\frac{3}{4}}{(1-t)}^{\frac{1}{2}}dt$$$$=\frac{\mathbf{\Gamma }\left(\frac{1}{4}\right)\mathbf{\Gamma }\left(\frac{3}{2}\right)}{2\sqrt{i}}$$

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