Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 86484 by Ar Brandon last updated on 28/Mar/20

$$\int \frac{x^6}{1+x^{12}}dx$$

Commented by Ar Brandon last updated on 29/Mar/20

Hi mathmax, I don't understand the second line of your solution. How did you do that ?

Commented by redmiiuser last updated on 29/Mar/20

$$sircanyouplscheck$$$$theanswer$$

Commented by Ar Brandon last updated on 29/Mar/20

How did you arrive there please ? ��

Commented by mathmax by abdo last updated on 29/Mar/20

$$z^{12}+1=0\Rightarrow z^{12}=-1=e^{i(2k+1)\pi }\Rightarrow z_k=e^{\frac{i(2k+1)\pi }{12}}k\in [[0,11]$$$$\Rightarrow \frac{x^6}{x^{12}+1}=\frac{x^6}{\prod _{k=0}^{11}(x-z_k)}=\sum _{k=0}^{11}\frac{a_k}{x-z_k}$$$$a_k=\frac{z_k^6}{12z_k^{11}}=\frac{z_k^7}{-12}\Rightarrow \frac{x^6}{x^{12}+1}=-\frac{1}{12}\sum _{k=0}^{11}\frac{z_k^7}{x-z_k}\Rightarrow$$$$\int \frac{x^6}{x^{12}+1}dx=-\frac{1}{12}\sum _{k=0}^{11}{z}_k^{7}ln(x-z_k)+C$$

Commented by mathmax by abdo last updated on 29/Mar/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 Ar Brandon last updated on 29/Mar/20

Oh ! Thanks ��

Commented by Ar Brandon last updated on 29/Mar/20

$$\mathrm{what}\mathrm{about}{\mathrm{a}}_{\mathrm{k}}=\frac{{\mathrm{z}}_{\mathrm{k}}^6}{{12\mathrm{z}}_{\mathrm{k}}^{11}}=\frac{{\mathrm{z}}_{\mathrm{k}}^7}{-12}?$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$because{z}_k^{12}=-1$$

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

$$\mathrm{hey}\mathrm{mathmax}\mathrm{where}\mathrm{i}\mathrm{can}\mathrm{see}\mathrm{the}$$$$\mathrm{demonstration}\mathrm{of}\mathrm{this}\mathrm{theorem}?$$

Answered by redmiiuser last updated on 29/Mar/20

$${(1+x^{12})}^{-1}$$$$=1+(-1)x^{12}+\frac{(-1)(-1-1)x^{24}}{2!}+\frac{(-1)(-1-1)(-1-2)x^{36}}{3!}+\frac{(-1)(-1-1)(-1-2)(-1-3)x^{48}}{4!}+....\mathrm{\infty }$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n.n!.{\left(x^{12}\right)}^n}{n!}$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n.{\left(x^{12}\right)}^n$$$$\therefore x^6.{(1+x^{12})}^{-1}$$$$=x^6.\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n.\left(x^{12n}\right)$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n.\left(x^{12n+6}\right)$$$$\therefore \int \underset{n=0}{\sum ^{\mathrm{\infty }}}{(-1)}^n.\left(x^{12n+6}\right).dx$$$$=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n.\left(x^{12n+7}\right)}{12n+7}$$

Commented by redmiiuser last updated on 29/Mar/20

$$welcomesir$$

Commented by Ar Brandon last updated on 29/Mar/20

Wow ! great idea. Thanks ! ��

Commented by redmiiuser last updated on 29/Mar/20

$$plschecktheanswer$$

Answered by MJS last updated on 29/Mar/20

$$\int \frac{x^6}{x^{12}+1}dx=$$$$=\frac{\sqrt{6}-\sqrt{2}}{24}\int \frac{x}{x^2-\frac{\sqrt{6}+\sqrt{2}}{2}x+1}dx-$$$$-\frac{\sqrt{6}-\sqrt{2}}{24}\int \frac{x}{x^2+\frac{\sqrt{6}+\sqrt{2}}{2}x+1}dx+$$$$+\frac{\sqrt{6}+\sqrt{2}}{24}\int \frac{x}{x^2-\frac{\sqrt{6}-\sqrt{2}}{2}x+1}dx-$$$$-\frac{\sqrt{6}+\sqrt{2}}{24}\int \frac{x}{x^2+\frac{\sqrt{6}-\sqrt{2}}{2}x+1}dx-$$$$-\frac{\sqrt{2}}{12}\int \frac{x}{x^2-\sqrt{2}x+1}dx+$$$$+\frac{\sqrt{2}}{12}\int \frac{x}{x^2+\sqrt{2}x+1}dx$$$$\mathrm{now}\mathrm{use}\mathrm{formula}$$

Commented by Ar Brandon last updated on 29/Mar/20

$$Ialsousedamethodsimilartoyoursanditwassobulky$$$$Imadeuseofthefactthat\mathit{c}\mathit{o}\mathit{s}\left(\frac{\mathit{\pi }}{12}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}\mathbf{and}\mathbf{sin}\left(\frac{\mathit{\pi }}{12}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}$$$$Wasthatthesameideawhichyouused?$$

Commented by MJS last updated on 29/Mar/20

$$\mathrm{yes}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com