can solve ∫ (dx/(x^(17) −1)) via elementary calculus?


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Question Number 77422 by john santu last updated on 06/Jan/20

$can solve \int \frac{dx}{x^{17} - 1} via$ $elementary calculus ?$
Commented by aliesam last updated on 07/Jan/20

$\int \frac{1}{x^{17} - 1} d x$ $= - \int \frac{1}{1 - x^{17}} (\frac{x}{x^{17}}) (\frac{x^{17}}{x}) d x$ $= - \int {(1 - x)}^{17 - 1} {(x)}^{1 - 17} {(x)}^{17 - 1} d x$ $= - \int {(1 - x)}^{17 - 1} {(x^{17})}^{\frac{1}{17} - 1} {(x)}^{17 - 1} d x$ $= - x_{2} F_{1} (1, \frac{1}{17}; \frac{18}{17}; x^{17}) + c$
	Answered by mind is power last updated on 06/Jan/20
	$X^(17) −1=Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) ) ∫(dx/(x^(17) −1))=∫(dx/(Π_(k=0) ^(16) (x−e^((2ikπ)/(17)) )))=∫{Σ_(k=0) ^(16) (a_k /(x−e^((2ikπ)/(17)) ))}dx a_k =(1/(17e^(((2ikπ)/(17)).16) ))=((e^(−((2ikπ)/(17))) )/(17)) ⇒∫(dx/(x^(17) −1))=∫Σ_(k=0) ^(16) (e^((−2ikπ)/(17)) /(17(x−e^((2ikπ)/(17)) ))).dx we can see that Σ_(k=0) ^(16) (e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))=((1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((−2iπ(17−k))/(17)) /(x−e^((2i(17−k)π)/(17)) )))) =(1/(x−1))+Σ_(k=1) ^8 ((e^(−((2ikπ)/(17))) /(x−e^((2ikπ)/(17)) ))+(e^((2ikπ)/(17)) /(x−e^((−2ikπ)/(17)) )))=(1/(x−1))+Σ_(k=1) ^8 ((2(xcos(((2kπ)/(17)))−cos(((2kπ)/(17)))))/(x^2 −2cos(((2πk)/(17)))x+1)) ∫2((xcos(((2kπ)/(17)))−cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))dx=∫{cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))}dx since ∣cos(((2kπ)/(17)))∣<1,∀k∈{1,.....8} ∫cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2xcos(((2πk)/(17)))+1))=cos(((2kπ)/(17)))ln(x^2 −2xcos(((2kπ)/(17)))x+1) ∫((2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))=(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))∫(dx/((x−cos(((2kπ)/(17)))^2 +sin^2 (((2kπ)/(17))))) =(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17)))) we Get ∫(dx/(x^(17) −1))=(1/(17))∫{(1/(x−1))+Σ_(k=1) ^8 cos(((2kπ)/(17)))((2x−2cos(((2kπ)/(17))))/(x^2 −2cos(((2kπ)/(17)))x+1))+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17)))).(1/(x^2 −2xcos(((2kπ)/(17)))+1))}dx =(1/(17)){ln∣x−1∣+Σ_(k=1) ^8 [cos(((2kπ)/(17)))ln(x^2 −2xcos(((2πk)/(17)))+1)+(2cos^2 (((2kπ)/(17)))−2cos(((2kπ)/(17))))arctan((x/(sin(((2kπ)/(17)))))−cot(((2kπ)/(17))))]}+c c∈R constant$
	$X^{17} - 1 = \underset{k = 0}{\prod^{16}} (x - e^{\frac{2 ik π}{17}})$ $\int \frac{dx}{x^{17} - 1} = \int \frac{dx}{\underset{k = 0}{\prod^{16}} (x - e^{\frac{2 ik π}{17}})} = \int {\underset{k = 0}{\sum^{16}} \frac{a_{k}}{x - e^{\frac{2 ik π}{17}}}} dx$ $a_{k} = \frac{1}{{17 e}^{\frac{2 ik π}{17} . 16}} = \frac{e^{- \frac{2 ik π}{17}}}{17}$ $\Rightarrow \int \frac{dx}{x^{17} - 1} = \int \underset{k = 0}{\sum^{16}} \frac{e^{\frac{- 2 ik π}{17}}}{17 (x - e^{\frac{2 ik π}{17}})} . dx$ $we can see that$ $\underset{k = 0}{\sum^{16}} \frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} = (\frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} (\frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} + \frac{e^{\frac{- 2 i π (17 - k)}{17}}}{x - e^{\frac{2 i (17 - k) π}{17}}}))$ $= \frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} (\frac{e^{- \frac{2 ik π}{17}}}{x - e^{\frac{2 ik π}{17}}} + \frac{e^{\frac{2 ik π}{17}}}{x - e^{\frac{- 2 ik π}{17}}}) = \frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} \frac{2 (xcos (\frac{2 k π}{17}) - \cos (\frac{2 k π}{17}))}{x^{2} - 2 \cos (\frac{2 π k}{17}) x + 1}$ $\int 2 \frac{xcos (\frac{2 k π}{17}) - \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} dx = \int {\cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} + \frac{{2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1}} dx$ $since ∣ \cos (\frac{2 k π}{17}) ∣< 1, \forall k \in {1, . . . . . 8}$ $\int \cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 xcos (\frac{2 π k}{17}) + 1} = \cos (\frac{2 k π}{17}) \ln (x^{2} - 2 xcos (\frac{2 k π}{17}) x + 1)$ $\int \frac{{2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} = ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \int \frac{dx}{(x - \cos {(\frac{2 k π}{17})}^{2} + \sin^{2} (\frac{2 k π}{17})}$ $= ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \arctan (\frac{x}{\sin (\frac{2 k π}{17})} - \cot (\frac{2 k π}{17}))$ $we Get$ $\int \frac{dx}{x^{17} - 1} = \frac{1}{17} \int {\frac{1}{x - 1} + \underset{k = 1}{\sum^{8}} \cos (\frac{2 k π}{17}) \frac{2 x - 2 \cos (\frac{2 k π}{17})}{x^{2} - 2 \cos (\frac{2 k π}{17}) x + 1} + ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) . \frac{1}{x^{2} - 2 xcos (\frac{2 k π}{17}) + 1}} dx$ $= \frac{1}{17} {\ln ∣ x - 1 ∣ + \underset{k = 1}{\sum^{8}} [\cos (\frac{2 k π}{17}) \ln (x^{2} - 2 xcos (\frac{2 π k}{17}) + 1) + ({2 \cos}^{2} (\frac{2 k π}{17}) - 2 \cos (\frac{2 k π}{17})) \arctan (\frac{x}{\sin (\frac{2 k π}{17})} - \cot (\frac{2 k π}{17}))]} + c$ $c \in R constant$
	Commented by john santu last updated on 06/Jan/20

	$thanks you sir$
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