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Question Number 102470 by  M±th+et+s last updated on 09/Jul/20

$$\int \frac{dx}{x^{10}+x^2}$$$$$$

Answered by Ar Brandon last updated on 09/Jul/20

$$\mathcal{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^{10}+{\mathrm{x}}^2}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2({\mathrm{x}}^8+1)}$$$$\frac{1}{{\mathrm{x}}^2({\mathrm{x}}^8+1)}=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2}+\frac{{\mathrm{cx}}^7+{\mathrm{dx}}^6+{\mathrm{ex}}^5+{\mathrm{fx}}^4+{\mathrm{gx}}^3+{\mathrm{hx}}^2+\mathrm{ix}+\mathrm{j}}{{\mathrm{x}}^8+1}$$$$\mathrm{x}=0\Rightarrow \mathrm{b}=1,\mathrm{a}+\mathrm{c}=0,\mathrm{d}+\mathrm{b}=0\Rightarrow \mathrm{d}=-1$$$$\mathrm{e}=0,\mathrm{f}=0,\mathrm{g}=0,\mathrm{h}=0,\mathrm{i}=0,\mathrm{j}=0,\mathrm{a}=0\Rightarrow \mathrm{c}=0$$$$\frac{1}{{\mathrm{x}}^2({\mathrm{x}}^8+1)}=\frac{1}{{\mathrm{x}}^2}-\frac{{\mathrm{x}}^6}{{\mathrm{x}}^8+1}$$$$\Rightarrow \mathcal{I}=\int \{\frac{1}{{\mathrm{x}}^2}-\frac{{\mathrm{x}}^6}{{\mathrm{x}}^8+1}\}\mathrm{dx}=-\frac{1}{\mathrm{x}}-\int \frac{{\mathrm{x}}^6}{{\mathrm{x}}^8+1}\mathrm{dx}$$$$\mathcal{J}=\int \frac{{\mathrm{x}}^6}{{\mathrm{x}}^8+1}\mathrm{dx},{\mathrm{x}}^8+1=0\Rightarrow {\mathrm{x}}^8={\mathrm{e}}^{(2\mathrm{k}+1)\pi \mathrm{i}}\Rightarrow {\mathrm{x}}_{\mathrm{k}}={\mathrm{e}}^{\frac{(2\mathrm{k}+1)\pi \mathrm{i}}{8}}$$$$\Rightarrow {\mathrm{x}}^8+1=\underset{\mathrm{k}=0}{\prod ^7}(\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})$$$$\Rightarrow \mathcal{J}=\int \frac{{\mathrm{x}}^6}{\underset{\mathrm{k}=0}{\prod ^7}(\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})}\mathrm{dx}=\int \underset{\mathrm{k}=0}{\sum ^7}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{x}-{\mathrm{x}}_{\mathrm{k}}}\mathrm{dx}$$$${\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{x}}_{\mathrm{k}}^6}{{8\mathrm{x}}_{\mathrm{k}}^7}=\frac{{\mathrm{x}}_{\mathrm{k}}^7}{-8}\Rightarrow \mathcal{J}=-\frac{1}{8}\int \underset{\mathrm{k}=0}{\sum ^7}\frac{{\mathrm{x}}_{\mathrm{k}}^7\mathrm{dx}}{(\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})}$$$$\Rightarrow \mathcal{J}=-\frac{1}{8}\underset{\mathrm{k}=0}{\sum ^7}{\mathrm{x}}_{\mathrm{k}}^7\ln (\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})$$$$\Rightarrow \int \frac{\mathrm{dx}}{{\mathrm{x}}^{10}+{\mathrm{x}}^2}=-\frac{1}{\mathrm{x}}+\frac{1}{8}\underset{\mathrm{k}=0}{\sum ^7}{\mathrm{x}}_{\mathrm{k}}^7\ln (\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})+\mathcal{C}$$$$$$$$Mr{Mathmax}^{\prime }\mathrm{s}favoritemethod.Haha!$$$$Ihope{I}^{\prime }vedoneitintherightway.$$

Commented by  M±th+et+s last updated on 09/Jul/20

$$welldonesir$$

Commented by Ar Brandon last updated on 09/Jul/20

��

Commented by mathmax by abdo last updated on 09/Jul/20

$$\mathrm{yes}\mathrm{your}\mathrm{answer}\mathrm{is}\mathrm{correct}$$

Commented by floor(10²Eta[1]) last updated on 09/Jul/20

$$\mathrm{can}\mathrm{you}\mathrm{explain}\mathrm{me}\mathrm{how}\mathrm{did}\mathrm{you}\mathrm{go}\mathrm{from}$$$$\mathrm{here}:\int \frac{{\mathrm{x}}^6}{\underset{\mathrm{k}=0}{\prod ^7}(\mathrm{x}-{\mathrm{x}}_{\mathrm{k}})}\mathrm{dx}\mathrm{to}:=\int \underset{\mathrm{k}=0}{\sum ^7}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{x}-{\mathrm{x}}_{\mathrm{k}}}\mathrm{dx}$$$$\mathrm{and}\mathrm{why}{\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{x}}_{\mathrm{k}}^6}{{8\mathrm{x}}_{\mathrm{k}}^7}$$

Commented by Ar Brandon last updated on 09/Jul/20

$$\mathrm{Hi}!\mathrm{Mathmax}\mathrm{explained}\mathrm{this}\mathrm{to}\mathrm{me}\mathrm{in}\mathrm{Q}86484.$$$$\mathrm{I}\mathrm{guess}\mathrm{you}\mathrm{have}\mathrm{a}\mathrm{look}\mathrm{at}\mathrm{it}.$$

Commented by Ar Brandon last updated on 09/Jul/20

$$$$$$itsaliketheoremifF=\frac{p\left(x\right)}{Q\left(x\right)}withdegp<degQand$$$$Qwithoutrealroots(\Rightarrow Q\left(x\right)=\lambda \prod _i(x-z_i))so$$$$F\left(x\right)=\sum _i\frac{a_i}{x-z_i}and{a}_i=\frac{p\left(z_i\right)}{Q^{{}^{\prime }}\left(z_i\right)}$$

Commented by floor(10²Eta[1]) last updated on 09/Jul/20

$$\mathrm{thanks}!{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{really}\mathrm{nice},\mathrm{do}\mathrm{you}\mathrm{know}\mathrm{where}$$$$\mathrm{i}\mathrm{can}\mathrm{see}\mathrm{the}\mathrm{demonstration}\mathrm{of}\mathrm{this}?$$

Commented by Ar Brandon last updated on 09/Jul/20

Thanks for your approval Sir Mathmax. ��

Commented by Ar Brandon last updated on 09/Jul/20

No idea, I don't know where you can find this demo. Maybe a little search on Google might help.��

Commented by mathmax by abdo last updated on 10/Jul/20

$$\mathrm{you}\mathrm{are}\mathrm{welcome}\mathrm{sir}$$

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