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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[]the[]integral[]of[]dt(1+t10)

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

11+x=1+n1nk=1(12k)n!xn=1+n1(1)n(2n)!4n.(n!)2xn11+t10=1+n1(1)n.2n!4n.n!.n!.x10ndt1+t10=x(1+n1(1)n(2n)!4n.(n!)2.x10n10n+1)+c2n!=4n.n!.n1k=0(k+12)=x(1+n1(1)n.4n.n!.n1k=0(k+12)4n.n!.10(n+110).x10nn!)+c=x(1+n1n1k=0(k+12).n1k=0(k+110)n1k=0(k+1110).(x10)nn!)+c=x2F1(12,110;1110;x10)+c

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

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

welldone.icheckedititsniceandcorrect

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

thankyouSiricantfindyourpostnmΓ(n+12).Γ(m+12)Γ(n)Γ(m)......canyousendtheIdofquationigotanideathankyou

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

go to q 98434

gotoq98434

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

thank you so much for solving my problems

thankyousomuchforsolvingmyproblems

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

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

withePleasurthankyouiwillpostsolutionnotyetfinishthisone

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