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

find ∫_0 ^1 (x^2 /( (√(x^4 +1))))dx

$$\mathrm{find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{1}}}\mathrm{dx} \\ $$

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

(1/( (√(1+x))))=Σ_(n=0) ^∞ ((((1/2))_n )/(n!))(−1)^n x^n (x^2 /( (√(1+x^4 ))))=Σ_(n=0) ^∞ ((((1/2))_n )/(n!))(−1)^n x^(4n+2) ∫_0 ^1 (x^2 /( (√(1+x^4 ))))=Σ_(n=0) ^∞ ((((1/2))_n )/(n!(4n+3)))(−1)^n =(1/3)Σ_(n=0) ^∞ ((((1/2))_n ((3/4))_n )/(((7/4))_n n!))(−1)^n As ,(1/(4n+3))=(1/4)((1/(n+(3/4))))=(1/4)(((Γ(n+(3/4)))/(Γ(n+(7/4)))))=(4/3).(1/4).((((3/4))_n )/(((7/4))_n )) =(1/3) _2 F_1 ((1/2),(3/4);(7/4);−1)

$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\left(−\mathrm{1}\right)^{{n}} {x}^{{n}} \\ $$$$\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{4}{n}+\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{4}{n}+\mathrm{3}\right)}\left(−\mathrm{1}\right)^{{n}} =\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)_{{n}} }{\left(\frac{\mathrm{7}}{\mathrm{4}}\right)_{{n}} {n}!}\left(−\mathrm{1}\right)^{{n}} \\ $$$${As}\:,\frac{\mathrm{1}}{\mathrm{4}{n}+\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{n}+\frac{\mathrm{3}}{\mathrm{4}}}\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\Gamma\left({n}+\frac{\mathrm{3}}{\mathrm{4}}\right)}{\Gamma\left({n}+\frac{\mathrm{7}}{\mathrm{4}}\right)}\right)=\frac{\mathrm{4}}{\mathrm{3}}.\frac{\mathrm{1}}{\mathrm{4}}.\frac{\left(\frac{\mathrm{3}}{\mathrm{4}}\right)_{{n}} }{\left(\frac{\mathrm{7}}{\mathrm{4}}\right)_{{n}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{4}};\frac{\mathrm{7}}{\mathrm{4}};−\mathrm{1}\right) \\ $$

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

thank you sir.

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

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