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Question Number 40684 by vajpaithegrate@gmail.com last updated on 26/Jul/18

$$\int \frac{{\mathrm{x}}^7-1}{\mathrm{logx}}\mathrm{dx}$$

Commented by math khazana by abdo last updated on 26/Jul/18

$$letI=\int \frac{x^7-1}{ln\left(x\right)}dxchangementln\left(x\right)=tgive$$$$I=\int \frac{e^{7t}-1}{t}{e}^{t}dt$$$$=\int \frac{e^{8t}-e^t}{t}dtbut{e}^{8t}=\sum _{n=0}^{\mathrm{\infty }}\frac{{\left(8t\right)}^n}{n!}and$$$$e^t=\sum _{n=0}^{\mathrm{\infty }}\frac{t^n}{n!}\Rightarrow e^{8t}-e^t=\sum _{n=0}^{\mathrm{\infty }}\frac{{(8^n-1)}^{}{t}^n}{n!}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{8^n-1}{n!}{t}^n\Rightarrow I=\int \sum _{n=1}^{\mathrm{\infty }}\frac{8^n-1}{n!}{t}^{n-1}$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{8^n-1}{n!}\frac{1}{n}{t}^n\Rightarrow$$$$I=\sum _{n=1}^{\mathrm{\infty }}\frac{8^n-1}{n(n!)}{\left(ln\left(x\right)\right)}^n.$$

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