Integration Questions

Question Number 162238 by amin96 last updated on 27/Dec/21

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{7}} +\mathrm{1}}\boldsymbol{\mathrm{dx}}=? \\$$

Answered by Ar Brandon last updated on 27/Dec/21

$${z}^{\mathrm{7}} +\mathrm{1}=\mathrm{0}\Rightarrow{z}_{{k}} ={e}^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{7}}{i}\pi} ,\:{k}\in\left[\mathrm{0},\:\mathrm{6}\right],\:{k}\in\mathbb{Z} \\$$ $${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{7}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dz}}{\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\prod}}\left({z}−{z}_{{k}} \right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\frac{{a}_{{k}} }{{z}−{z}_{{k}} }{dz} \\$$ $${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{7}{z}_{{k}} ^{\mathrm{6}} }=−\frac{{z}_{{k}} }{\mathrm{7}} \\$$ $${I}=−\frac{\mathrm{1}}{\mathrm{7}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}_{{k}} }{{z}−{z}_{{k}} }{dz}=−\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\left[\frac{{z}_{{k}} \mathrm{ln}\mid{z}−{z}_{{k}} \mid}{\mathrm{7}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\$$