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Question Number 220037 by Nicholas666 last updated on 04/May/25
∫ (1/( (√(x^8 + 1)))) dx
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx} \\ $$$$ \\ $$
Answered by MrGaster last updated on 04/May/25
∫ (1/( (√(x^8 + 1)))) dx=∫Σ_(n=0) ^∞ (((−1)^n ((1/2))_n )/(n!))x^(8n) dx=Σ_(n=0) ^∞ (((−1)^n ((1/2))_n )/(n!(8n+1)))x^(8n+1) +C=xΣ_(n=0) ^∞ ((((1/8))_n ((1/2))_n )/(((9/8))_n n!))(−x^8 )^n +C=x _2 F_1 ((1/8),(1/2),(9/8);−x^8 )+C (2):∫ (1/( (√(x^8 + 1)))) dx=∫(1+x^8 )^(−1/2) dx=∫Σ_(n=0) ^∞ (((−1/2)_n )/(n!))x^(8n) dx=Σ_(n=0) ^∞ (((−1/2)_n )/(n!))∫x^(8n) dx=Σ_(n=0) ^∞ (((−1/2)_n )/(n!))((x^(8n+1) /(8n+1)))+C=xΣ_(n=0) ^∞ (((−1/2)_n (1/8)_n )/((9/8)_n n!))(−x^8 )^n =x _2 F_1 ((1/8),(1/2),(9/8);−x^8 )+C
$$\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!}{x}^{\mathrm{8}{n}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{{n}!\left(\mathrm{8}{n}+\mathrm{1}\right)}{x}^{\mathrm{8}{n}+\mathrm{1}} +{C}={x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\frac{\mathrm{1}}{\mathrm{8}}\right)_{{n}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)_{{n}} }{\left(\frac{\mathrm{9}}{\mathrm{8}}\right)_{{n}} {n}!}\left(−{x}^{\mathrm{8}} \right)^{{n}} +{C}={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{8}};−{x}^{\mathrm{8}} \right)+{C} \\ $$$$\left(\mathrm{2}\right):\int\:\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{8}} \:+\:\mathrm{1}}}\:{dx}=\int\left(\mathrm{1}+{x}^{\mathrm{8}} \right)^{−\mathrm{1}/\mathrm{2}} {dx}=\int\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}/\mathrm{2}\right)_{{n}} }{{n}!}{x}^{\mathrm{8}{n}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}/\mathrm{2}\right)_{{n}} }{{n}!}\int{x}^{\mathrm{8}{n}} {dx}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}/\mathrm{2}\right)_{{n}} }{{n}!}\left(\frac{{x}^{\mathrm{8}{n}+\mathrm{1}} }{\mathrm{8}{n}+\mathrm{1}}\right)+{C}={x}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}/\mathrm{2}\right)_{{n}} \left(\mathrm{1}/\mathrm{8}\right)_{{n}} }{\left(\mathrm{9}/\mathrm{8}\right)_{{n}} {n}!}\left(−{x}^{\mathrm{8}} \right)^{{n}} ={x}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{8}},\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{9}}{\mathrm{8}};−{x}^{\mathrm{8}} \right)+{C} \\ $$

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