Question Number 158492 by bobhans last updated on 05/Nov/21
$$\:\int\:\frac{\mathrm{dx}}{\mathrm{1}+\mathrm{x}^{\mathrm{7}} }\:? \\ $$
Answered by Ar Brandon last updated on 05/Nov/21
![x^7 +1=0⇒x^7 =e^((2k+1)iπ) ⇒x_k =e^((((2k+1)/7))iπ) , k∈[0, 6] k∈Z I=∫(dx/(x^7 +1))=∫(dx/(Π_(k=0) ^6 (x−x_k )))=∫Σ_(k=0) ^6 (a_k /(x−x_k ))dx , a_k =(1/(7x_k ^6 ))=−(x_k /7) I=−(1/7)∫Σ_(k=0) ^6 (x_k /(x−x_k ))dx=−(1/7)Σ_(k=0) ^6 x_k ln∣x−x_k ∣+C](https://www.tinkutara.com/question/Q158513.png)
$${x}^{\mathrm{7}} +\mathrm{1}=\mathrm{0}\Rightarrow{x}^{\mathrm{7}} ={e}^{\left(\mathrm{2}{k}+\mathrm{1}\right){i}\pi} \Rightarrow{x}_{{k}} ={e}^{\left(\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{7}}\right){i}\pi} ,\:{k}\in\left[\mathrm{0},\:\mathrm{6}\right]\:{k}\in\mathbb{Z} \\ $$$${I}=\int\frac{{dx}}{{x}^{\mathrm{7}} +\mathrm{1}}=\int\frac{{dx}}{\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\prod}}\left({x}−{x}_{{k}} \right)}=\int\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\frac{{a}_{{k}} }{{x}−{x}_{{k}} }{dx}\:,\:{a}_{{k}} =\frac{\mathrm{1}}{\mathrm{7}{x}_{{k}} ^{\mathrm{6}} }=−\frac{{x}_{{k}} }{\mathrm{7}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{7}}\int\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\frac{{x}_{{k}} }{{x}−{x}_{{k}} }{dx}=−\frac{\mathrm{1}}{\mathrm{7}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}{x}_{{k}} \mathrm{ln}\mid{x}−{x}_{{k}} \mid+{C} \\ $$