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Question Number 77422 by john santu last updated on 06/Jan/20
can solve ∫ (dx/(x^(17) −1)) via   elementary calculus?
cansolvedxx171viaelementarycalculus?
Commented by aliesam last updated on 07/Jan/20
∫(1/(x^(17) −1)) dx  = −∫(1/(1−x^(17) )) ((x/x^(17) ))((x^(17) /x)) dx  =−∫(1−x)^(17−1)  (x)^(1−17)  (x)^(17−1)  dx  =−∫(1−x)^(17−1)  (x^(17) )^((1/(17))−1) (x)^(17−1)  dx  =−x_2 F_1 (1, (1/(17)) ; ((18)/(17)) ;x^(17) )+c
1x171dx=11x17(xx17)(x17x)dx=(1x)171(x)117(x)171dx=(1x)171(x17)1171(x)171dx=x2F1(1,117;1817;x17)+c
Answered by mind is power last updated on 06/Jan/20
X^(17) −1=Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) )  ∫(dx/(x^(17) −1))=∫(dx/(Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) )))=∫{Σ_(k=0) ^(16) (a_k /(x−e^((2ikπ)/(17)) ))}dx  a_k =(1/(17e^(((2ikπ)/(17)).16) ))=((e^(−((2ikπ)/(17)))  )/(17))  ⇒∫(dx/(x^(17) −1))=∫Σ_(k=0) ^(16) (e^((−2ikπ)/(17)) /(17(x−e^((2ikπ)/(17)) ))).dx  we can see that  Σ_(k=0) ^(16) (e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))=((1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((−2iπ(17−k))/(17)) /(x−e^((2i(17−k)π)/(17)) ))))  =(1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((2ikπ)/(17)) /(x−e^((−2ikπ)/(17)) )))=(1/(x−1))+Σ_(k=1) ^8 ((2(xcos(((2kπ)/(17)))−cos(((2kπ)/(17)))))/(x^2 −2cos(((2πk)/(17)))x+1))  ∫2((xcos(((2kπ)/(17)))−cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))dx=∫{cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))}dx  since ∣cos(((2kπ)/(17)))∣<1,∀k∈{1,.....8}   ∫cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2xcos(((2πk)/(17)))+1))=cos(((2kπ)/(17)))ln(x^2 −2xcos(((2kπ)/(17)))x+1)  ∫((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))=(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))∫(dx/((x−cos(((2kπ)/(17)))^2 +sin^2 (((2kπ)/(17)))))  =(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17))))  we Get  ∫(dx/(x^(17) −1))=(1/(17))∫{(1/(x−1))+Σ_(k=1) ^8 cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17)))).(1/(x^2 −2xcos(((2kπ)/(17)))+1))}dx  =(1/(17)){ln∣x−1∣+Σ_(k=1) ^8 [cos(((2kπ)/(17)))ln(x^2 −2xcos(((2πk)/(17)))+1)+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17))))]}+c  c∈R constant
X171=16k=0(xe2ikπ17)dxx171=dx16k=0(xe2ikπ17)={16k=0akxe2ikπ17}dxak=117e2ikπ17.16=e2ikπ1717dxx171=16k=0e2ikπ1717(xe2ikπ17).dxwecanseethat16k=0e2ikπ17xe2ikπ17=(1x1+8k=1(e2ikπ17xe2ikπ17+e2iπ(17k)17xe2i(17k)π17))=1x1+8k=1(e2ikπ17xe2ikπ17+e2ikπ17xe2ikπ17)=1x1+8k=12(xcos(2kπ17)cos(2kπ17))x22cos(2πk17)x+12xcos(2kπ17)cos(2kπ17)x22cos(2kπ17)x+1dx={cos(2kπ17)2x2cos(2kπ17)x22cos(2kπ17)x+1+2cos2(2kπ17)2cos(2kπ17)x22cos(2kπ17)x+1}dxsincecos(2kπ17)∣<1,k{1,..8}cos(2kπ17)2x2cos(2kπ17)x22xcos(2πk17)+1=cos(2kπ17)ln(x22xcos(2kπ17)x+1)2cos2(2kπ17)2cos(2kπ17)x22cos(2kπ17)x+1=(2cos2(2kπ17)2cos(2kπ17))dx(xcos(2kπ17)2+sin2(2kπ17)=(2cos2(2kπ17)2cos(2kπ17))arctan(xsin(2kπ17)cot(2kπ17))weGetdxx171=117{1x1+8k=1cos(2kπ17)2x2cos(2kπ17)x22cos(2kπ17)x+1+(2cos2(2kπ17)2cos(2kπ17)).1x22xcos(2kπ17)+1}dx=117{lnx1+8k=1[cos(2kπ17)ln(x22xcos(2πk17)+1)+(2cos2(2kπ17)2cos(2kπ17))arctan(xsin(2kπ17)cot(2kπ17))]}+ccRconstant
Commented by john santu last updated on 06/Jan/20
thanks you sir
thanksyousir

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