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Question Number 119894 by danielasebhofoh last updated on 27/Oct/20

Answered by mathmax by abdo last updated on 28/Oct/20

$$\mathrm{let}{\mathrm{I}}_{\mathrm{n}}=\int \frac{\mathrm{dx}}{\mathrm{x}({\mathrm{x}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}})}\mathrm{if}\mathrm{a}\ne 0\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\mathrm{x}=\mathrm{at}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}=\int \frac{\mathrm{adt}}{\mathrm{at}({\mathrm{a}}^{\mathrm{n}}{\mathrm{t}}^{\mathrm{n}}-{\mathrm{a}}^{\mathrm{n}})}=\frac{1}{{\mathrm{a}}^{\mathrm{n}}}\int \frac{\mathrm{dt}}{\mathrm{t}({\mathrm{t}}^{\mathrm{n}}-1)}\mathrm{let}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{1}{\mathrm{t}({\mathrm{t}}^{\mathrm{n}}-1)}\mathrm{inside}\mathrm{C}\left(\mathrm{x}\right)$$$${\mathrm{t}}^{\mathrm{n}}-1=0{\Rightarrow }_{}{\mathrm{t}}_{\mathrm{k}}={\mathrm{e}}^{\mathrm{i}\frac{2\mathrm{k}\pi }{\mathrm{n}}}\mathrm{and}\mathrm{k}\in \left[[0,\mathrm{n}-1]\right]\Rightarrow$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{1}{\mathrm{t}\prod _{\mathrm{k}=0}^{\mathrm{n}-1}(\mathrm{t}-{\mathrm{t}}_{\mathrm{k}})}=\frac{\mathrm{c}}{\mathrm{t}}+\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{t}-{\mathrm{t}}_{\mathrm{k}}}$$$$\mathrm{c}=-1\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=-\frac{1}{\mathrm{t}}+\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{t}-{\mathrm{t}}_{\mathrm{k}}}\Rightarrow$$$$\mathrm{F}\left(\mathrm{t}\right)+\frac{1}{\mathrm{t}}=\frac{1}{\mathrm{t}({\mathrm{t}}^{\mathrm{n}}-1)}+\frac{1}{\mathrm{t}}=\frac{1}{\mathrm{t}}\{\frac{1}{{\mathrm{t}}^{\mathrm{n}}-1}+1\}=\frac{1}{\mathrm{t}}\times \frac{{\mathrm{t}}^{\mathrm{n}}}{{\mathrm{t}}^{\mathrm{n}}-1}=\frac{{\mathrm{t}}^{\mathrm{n}-1}}{{\mathrm{t}}^{\mathrm{n}}-1}$$$$\Rightarrow {\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{t}}_{\mathrm{k}}^{\mathrm{n}-1}}{\mathrm{n}{\mathrm{t}}_{\mathrm{k}}^{\mathrm{n}-1}}=\frac{1}{\mathrm{n}}\Rightarrow \mathrm{F}\left(\mathrm{t}\right)=-\frac{1}{\mathrm{t}}+\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\frac{1}{\mathrm{t}-{\mathrm{t}}_{\mathrm{k}}}\Rightarrow$$$$\int \mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=-\ln \mid \mathrm{t}\mid +\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\ln (\mathrm{t}-{\mathrm{e}}^{\frac{\mathrm{i}2\mathrm{k}\pi }{\mathrm{n}}})+\mathrm{C}\Rightarrow$$$${\mathrm{I}}_{\mathrm{n}}=\frac{1}{{\mathrm{a}}^{\mathrm{n}}}\{-\ln \mid \frac{\mathrm{x}}{\mathrm{a}}\mid +\frac{1}{\mathrm{n}}\sum _{\mathrm{k}=0}^{\mathrm{n}-1}\ln (\frac{\mathrm{x}}{\mathrm{a}}-{\mathrm{e}}^{\frac{\mathrm{i}2\mathrm{k}\pi }{\mathrm{n}}})\}+\mathrm{C}$$

Answered by TANMAY PANACEA last updated on 28/Oct/20

$$simpleproblem$$$$\int \frac{x^{n-1}dx}{x^n(x^n-a^n)}\left[\mathit{m}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{p}\mathit{l}\mathit{y}{\mathit{N}}_{r}and{D}_{r}by{x}^{n-1}\right]$$$$t=x^n\to \frac{dt}{dx}={nx}^{n-1}$$$$\int \frac{dt}{n\left(t\right)(t-a^n)}$$$$\frac{1}{{na}^n}\int (\frac{1}{t-a^n}-\frac{1}{t})dt$$$$\frac{1}{{na}^n}ln\left(\frac{t-a^n}{t}\right)+c\to \frac{1}{{\mathit{n}\mathit{a}}^{\mathit{n}}}\mathit{l}\mathit{n}\left(\frac{{\mathit{x}}^{\mathit{n}}-{\mathit{a}}^{\mathit{n}}}{{\mathit{x}}^{\mathit{n}}}\right)+\mathit{c}$$

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