Question Number 84497 by M±th+et£s last updated on 13/Mar/20
$$\int\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{5}} }\:{dx} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 13/Mar/20
![z^5 +1 =0 ⇒z^5 =−1 let z=re^(iθ) ⇒r^5 e^(i5θ) =e^(i(2k+1)π) ⇒ r=1 and θ_k =(((2k+1)π)/5) ⇒z_k =e^(i(((2k+1)π)/5)) k∈[[0,4]] (z^2 /(z^5 +1)) =(z^2 /(Π_(k=0) ^4 (z−z_k ))) =Σ_(k=0) ^4 (a_k /(z−z_k )) a_k =(z_k ^2 /(5z_4 ^4 )) =(z_k ^3 /(−5)) =−(1/5)z_k ^3 ⇒∫ (x^2 /(x^5 +1))dx =−(1/5)∫ Σ_(k=0) ^4 (z_k ^3 /(x−z_k ))dx =−(1/5)Σ_(k=0) ^4 z_k ^3 ln(x−z_k ) +C](https://www.tinkutara.com/question/Q84523.png)
$${z}^{\mathrm{5}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow{z}^{\mathrm{5}} =−\mathrm{1}\:{let}\:{z}={re}^{{i}\theta} \:\Rightarrow{r}^{\mathrm{5}} \:{e}^{{i}\mathrm{5}\theta} ={e}^{{i}\left(\mathrm{2}{k}+\mathrm{1}\right)\pi} \:\Rightarrow \\ $$$${r}=\mathrm{1}\:{and}\:\theta_{{k}} =\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}}\:\Rightarrow{z}_{{k}} ={e}^{{i}\frac{\left(\mathrm{2}{k}+\mathrm{1}\right)\pi}{\mathrm{5}}} \:\:\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right] \\ $$$$\frac{{z}^{\mathrm{2}} }{{z}^{\mathrm{5}} \:+\mathrm{1}}\:=\frac{{z}^{\mathrm{2}} }{\prod_{{k}=\mathrm{0}} ^{\mathrm{4}} \left({z}−{z}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{{a}_{{k}} }{{z}−{z}_{{k}} } \\ $$$${a}_{{k}} =\frac{{z}_{{k}} ^{\mathrm{2}} }{\mathrm{5}{z}_{\mathrm{4}} ^{\mathrm{4}} }\:=\frac{{z}_{{k}} ^{\mathrm{3}} }{−\mathrm{5}}\:=−\frac{\mathrm{1}}{\mathrm{5}}{z}_{{k}} ^{\mathrm{3}} \:\Rightarrow\int\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{5}} \:+\mathrm{1}}{dx}\:=−\frac{\mathrm{1}}{\mathrm{5}}\int\:\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:\frac{{z}_{{k}} ^{\mathrm{3}} }{{x}−{z}_{{k}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}}\sum_{{k}=\mathrm{0}} ^{\mathrm{4}} \:{z}_{{k}} ^{\mathrm{3}} \:\:{ln}\left({x}−{z}_{{k}} \right)\:+{C} \\ $$