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Question Number 98929 by Dwaipayan Shikari last updated on 17/Jun/20

Find[]the[]integral[]of[] ∫(dt/(√((1+t^(10) ))))

$${Find}\left[\right]{the}\left[\right]{integral}\left[\right]{of}\left[\right] \\ $$$$ \\ $$$$\int\frac{{dt}}{\sqrt{\left(\mathrm{1}+{t}^{\mathrm{10}} \right)}} \\ $$

Commented by maths mind last updated on 17/Jun/20

(1/(√(1+x)))=1+Σ_(n≥1) ((Π_(k=1) ^n ((1/2)−k))/(n!))x^n =1+Σ_(n≥1) (((−1)^n (2n)!)/(4^n .(n!)^2 ))x^n (1/(√(1+t^(10) )))=1+Σ_(n≥1) (((−1)^n .2n!)/(4^n .n!.n!)).x^(10n) ∫(dt/(√(1+t^(10) )))=x(1+Σ_(n≥1) (((−1)^n (2n)!)/(4^n .(n!)^2 )).(x^(10n) /(10n+1)))+c 2n!=4^n .n!.Π_(k=0) ^(n−1) (k+(1/2)) =x(1+Σ_(n≥1) (((−1)^n .4^n .n!.Π_(k=0) ^(n−1) (k+(1/2)))/(4^n .n!.10(n+(1/(10))))).(x^(10n) /(n!)))+c =x(1+Σ_(n≥1) ((Π_(k=0) ^(n−1) (k+(1/2)).Π_(k=0) ^(n−1) (k+(1/(10))))/(Π_(k=0) ^(n−1) (k+((11)/(10))))).(((−x^(10) )^n )/(n!)))+c =x _2 F_1 ((1/2),(1/(10));((11)/(10));−x^(10) )+c

$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{x}}}=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\mathrm{1}}{\mathrm{2}}−{k}\right)}{{n}!}{x}^{{n}} \\ $$$$=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}\right)!}{\mathrm{4}^{{n}} .\left({n}!\right)^{\mathrm{2}} }{x}^{{n}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{t}^{\mathrm{10}} }}=\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .\mathrm{2}{n}!}{\mathrm{4}^{{n}} .{n}!.{n}!}.{x}^{\mathrm{10}{n}} \\ $$$$\int\frac{{dt}}{\sqrt{\mathrm{1}+{t}^{\mathrm{10}} }}={x}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} \left(\mathrm{2}{n}\right)!}{\mathrm{4}^{{n}} .\left({n}!\right)^{\mathrm{2}} }.\frac{{x}^{\mathrm{10}{n}} }{\mathrm{10}{n}+\mathrm{1}}\right)+{c} \\ $$$$\mathrm{2}{n}!=\mathrm{4}^{{n}} .{n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$={x}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{{n}} .\mathrm{4}^{{n}} .{n}!.\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\mathrm{4}^{{n}} .{n}!.\mathrm{10}\left({n}+\frac{\mathrm{1}}{\mathrm{10}}\right)}.\frac{{x}^{\mathrm{10}{n}} }{{n}!}\right)+{c} \\ $$$$={x}\left(\mathrm{1}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{2}}\right).\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{1}}{\mathrm{10}}\right)}{\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+\frac{\mathrm{11}}{\mathrm{10}}\right)}.\frac{\left(−{x}^{\mathrm{10}} \right)^{{n}} }{{n}!}\right)+{c} \\ $$$$={x}\:\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{10}};\frac{\mathrm{11}}{\mathrm{10}};−{x}^{\mathrm{10}} \right)+{c} \\ $$

Commented by  M±th+et+s last updated on 17/Jun/20

well done. i checked it it′s nice and correct

$${well}\:{done}.\:{i}\:{checked}\:{it}\:{it}'{s}\:{nice}\:{and}\:{correct} \\ $$

Commented by maths mind last updated on 18/Jun/20

thank you Sir i cant find your post Σ_n Σ_m ((Γ(n+(1/2)).Γ(m+(1/2)))/(Γ(n)Γ(m)))...... can you send the Id of quation i got an idea thank you

$${thank}\:{you}\:{Sir}\: \\ $$$${i}\:{cant}\:{find}\:{your}\:{post}\:\underset{{n}} {\sum}\underset{{m}} {\sum}\frac{\Gamma\left({n}+\frac{\mathrm{1}}{\mathrm{2}}\right).\Gamma\left({m}+\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)\Gamma\left({m}\right)}...... \\ $$$${can}\:{you}\:{send}\:{the}\:{Id}\:{of}\:{quation}\:{i}\:{got}\:{an}\:{idea}\:\:\:{thank}\:{you} \\ $$

Commented by  M±th+et+s last updated on 18/Jun/20

go to q 98434

$${go}\:{to}\:{q}\:\mathrm{98434} \\ $$

Commented by  M±th+et+s last updated on 18/Jun/20

thank you so much for solving my problems

$${thank}\:{you}\:{so}\:{much}\:{for}\:{solving}\:{my}\:{problems} \\ $$

Commented by maths mind last updated on 20/Jun/20

withe Pleasur thank you i will post solution not yet finish this one

$${withe}\:{Pleasur}\:{thank}\:{you}\:{i}\:{will}\:{post}\:{solution} \\ $$$${not}\:{yet}\:{finish}\:{this}\:{one} \\ $$

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