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x-7-1-logx-dx-




Question Number 40684 by vajpaithegrate@gmail.com last updated on 26/Jul/18
∫((x^7 −1)/(logx))dx
$$\int\frac{\mathrm{x}^{\mathrm{7}} −\mathrm{1}}{\mathrm{logx}}\mathrm{dx} \\ $$
Commented by math khazana by abdo last updated on 26/Jul/18
let I = ∫    ((x^7 −1)/(ln(x)))dx  changement ln(x)=t give  I  = ∫    ((e^(7t)  −1)/t) e^t dt  = ∫    ((e^(8t)  −e^t )/t) dt  but  e^(8t)  =Σ_(n=0) ^∞    (((8t)^n )/(n!)) and  e^t  =Σ_(n=0) ^∞   (t^n /(n!)) ⇒e^(8t)  −e^t  =Σ_(n=0) ^∞   (((8^n  −1)^ t^n )/(n!))  =Σ_(n=1) ^∞    ((8^n  −1)/(n!))t^n  ⇒ I  = ∫  Σ_(n=1) ^∞   ((8^n −1)/(n!)) t^(n−1)   =Σ_(n=1) ^∞     ((8^n −1)/(n!)) (1/n) t^n  ⇒  I  = Σ_(n=1) ^∞    ((8^n −1)/(n(n!))) (ln(x))^n  .
$${let}\:{I}\:=\:\int\:\:\:\:\frac{{x}^{\mathrm{7}} −\mathrm{1}}{{ln}\left({x}\right)}{dx}\:\:{changement}\:{ln}\left({x}\right)={t}\:{give} \\ $$$${I}\:\:=\:\int\:\:\:\:\frac{{e}^{\mathrm{7}{t}} \:−\mathrm{1}}{{t}}\:{e}^{{t}} {dt} \\ $$$$=\:\int\:\:\:\:\frac{{e}^{\mathrm{8}{t}} \:−{e}^{{t}} }{{t}}\:{dt}\:\:{but}\:\:{e}^{\mathrm{8}{t}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\mathrm{8}{t}\right)^{{n}} }{{n}!}\:{and} \\ $$$${e}^{{t}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{n}} }{{n}!}\:\Rightarrow{e}^{\mathrm{8}{t}} \:−{e}^{{t}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{8}^{{n}} \:−\mathrm{1}\right)^{} {t}^{{n}} }{{n}!} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{8}^{{n}} \:−\mathrm{1}}{{n}!}{t}^{{n}} \:\Rightarrow\:{I}\:\:=\:\int\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{8}^{{n}} −\mathrm{1}}{{n}!}\:{t}^{{n}−\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\:\frac{\mathrm{8}^{{n}} −\mathrm{1}}{{n}!}\:\frac{\mathrm{1}}{{n}}\:{t}^{{n}} \:\Rightarrow \\ $$$${I}\:\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\mathrm{8}^{{n}} −\mathrm{1}}{{n}\left({n}!\right)}\:\left({ln}\left({x}\right)\right)^{{n}} \:. \\ $$

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