Question Number 108073 by bemath last updated on 14/Aug/20
$$\:\:\:\:\frac{\triangleq\mathcal{B}{e}\mathcal{M}{ath}\triangleq}{\prec} \\ $$$$\:\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{8}} −\mathrm{1}}\:? \\ $$
Answered by john santu last updated on 14/Aug/20
![((□JS⊡)/≢) I= ∫ ((x dx)/(x^8 −1)) set x^2 = z ⇒ 2x dx = dz I= (1/2)∫ (dz/(z^4 −1)) = (1/2)∫ (dz/((z^2 −1)(z^2 +1))) I=(1/4)∫[ (1/(z^2 −1))−(1/(z^2 +1))] dz I=(1/8)∫( (1/(z−1))−(1/(z+1)))dz−(1/4)tan^(−1) (z)+c I=(1/8)ln ∣((z−1)/(z+1))∣−(1/4)tan^(−1) (z)+c I=(1/8)ln ∣((x^2 −1)/(x^2 +1))∣−(1/4)tan^(−1) (x^2 )+c](https://www.tinkutara.com/question/Q108074.png)
$$\:\:\:\:\frac{\Box\mathcal{JS}\boxdot}{≢} \\ $$$${I}=\:\int\:\frac{{x}\:{dx}}{{x}^{\mathrm{8}} −\mathrm{1}}\: \\ $$$${set}\:{x}^{\mathrm{2}} \:=\:{z}\:\Rightarrow\:\mathrm{2}{x}\:{dx}\:=\:{dz}\: \\ $$$${I}=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dz}}{{z}^{\mathrm{4}} −\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{dz}}{\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left({z}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}\int\left[\:\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} +\mathrm{1}}\right]\:{dz} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\int\left(\:\frac{\mathrm{1}}{{z}−\mathrm{1}}−\frac{\mathrm{1}}{{z}+\mathrm{1}}\right){dz}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} \left({z}\right)+{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} \left({z}\right)+{c} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{ln}\:\mid\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{tan}^{−\mathrm{1}} \left({x}^{\mathrm{2}} \right)+{c} \\ $$
Commented by bemath last updated on 14/Aug/20
$${waw}…{cooll}\: \\ $$
Answered by malwaan last updated on 14/Aug/20
![∫(x/(x^8 −1))dx=(1/2)∫ ((d(x^2 ))/((x^2 )^4 −1)) y=x^2 ⇒(1/2)∫(dy/(y^4 −1)) =(1/2)∫[(dy/((y^2 +1)(y+1)(y−1)))]dy =(1/2)∫[((ay+b)/(y^2 +1))+(c/(y+1))+(d/(y−1))]dy y=1⇒4d=1⇒d=(1/4) y=−1⇒−4c=1⇒c=−(1/4) y=0⇒−b−c+d=1 ⇒b=(1/2)−1=−1 y=2⇒(2a−(1/2))(3)(1)+(−(1/4))(5)(1) +((1/4))(5)(3)=1⇒6a=0 ⇒a=0 ∴(1/2)∫[−(1/2)(1/(y^2 +1))−(1/4)(1/(y+1))+(1/4)(1/(y−1))]dy =−(1/4)tan^(−1) x^2 −(1/8)ln∣x^2 +1∣ +(1/8)ln∣x^2 −1∣+ C](https://www.tinkutara.com/question/Q108078.png)
$$\int\frac{{x}}{{x}^{\mathrm{8}} −\mathrm{1}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{d}\left({x}^{\mathrm{2}} \right)}{\left({x}^{\mathrm{2}} \right)^{\mathrm{4}} −\mathrm{1}} \\ $$$${y}={x}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dy}}{{y}^{\mathrm{4}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{{dy}}{\left({y}^{\mathrm{2}} +\mathrm{1}\right)\left({y}+\mathrm{1}\right)\left({y}−\mathrm{1}\right)}\right]{dy} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{{ay}+{b}}{{y}^{\mathrm{2}} +\mathrm{1}}+\frac{{c}}{{y}+\mathrm{1}}+\frac{{d}}{{y}−\mathrm{1}}\right]{dy} \\ $$$${y}=\mathrm{1}\Rightarrow\mathrm{4}{d}=\mathrm{1}\Rightarrow{d}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=−\mathrm{1}\Rightarrow−\mathrm{4}{c}=\mathrm{1}\Rightarrow{c}=−\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${y}=\mathrm{0}\Rightarrow−{b}−{c}+{d}=\mathrm{1} \\ $$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}=−\mathrm{1} \\ $$$${y}=\mathrm{2}\Rightarrow\left(\mathrm{2}{a}−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{3}\right)\left(\mathrm{1}\right)+\left(−\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{5}\right)\left(\mathrm{1}\right) \\ $$$$+\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\left(\mathrm{5}\right)\left(\mathrm{3}\right)=\mathrm{1}\Rightarrow\mathrm{6}{a}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{0} \\ $$$$\therefore\frac{\mathrm{1}}{\mathrm{2}}\int\left[−\frac{\mathrm{1}}{\mathrm{2}}\frac{\mathrm{1}}{{y}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{4}}\frac{\mathrm{1}}{{y}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{4}}\frac{\mathrm{1}}{{y}−\mathrm{1}}\right]{dy} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}{tan}^{−\mathrm{1}} {x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{x}^{\mathrm{2}} +\mathrm{1}\mid \\ $$$$+\frac{\mathrm{1}}{\mathrm{8}}{ln}\mid{x}^{\mathrm{2}} −\mathrm{1}\mid+\:{C} \\ $$
Answered by mathmax by abdo last updated on 14/Aug/20
![complex method z^8 −1=0 ⇒z^8 =e^(i2kπ) ⇒z_k =e^((i2kπ)/8) =e^((ikπ)/4) k ∈[[0,7]] ⇒F(z) =(z/(z^8 −1)) =(z/(Π_(k=0) ^7 (z−z_k ))) =Σ_(k=0) ^7 (a_k /(z−z_k )) a_k =(z_k /(8z_k ^7 )) =(z_k ^2 /8) ⇒F(z) =(1/8)Σ_(k=0) ^7 (z_k ^2 /(z−z_k )) ⇒ ∫ F(z)dz =(1/8)Σ_(k=0) ^7 z_k ^2 ln(z−z_k ) +C](https://www.tinkutara.com/question/Q108091.png)
$$\mathrm{complex}\:\mathrm{method}\:\:\:\:\:\:\mathrm{z}^{\mathrm{8}} −\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{8}} \:=\mathrm{e}^{\mathrm{i2k}\pi} \:\Rightarrow\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\frac{\mathrm{i2k}\pi}{\mathrm{8}}} \:=\mathrm{e}^{\frac{\mathrm{ik}\pi}{\mathrm{4}}} \\ $$$$\mathrm{k}\:\in\left[\left[\mathrm{0},\mathrm{7}\right]\right]\:\Rightarrow\mathrm{F}\left(\mathrm{z}\right)\:=\frac{\mathrm{z}}{\mathrm{z}^{\mathrm{8}} −\mathrm{1}}\:=\frac{\mathrm{z}}{\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \left(\mathrm{z}−\mathrm{z}_{\mathrm{k}} \right)}\:=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{z}−\mathrm{z}_{\mathrm{k}} } \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{z}_{\mathrm{k}} }{\mathrm{8z}_{\mathrm{k}} ^{\mathrm{7}} }\:=\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} }{\mathrm{8}}\:\Rightarrow\mathrm{F}\left(\mathrm{z}\right)\:=\frac{\mathrm{1}}{\mathrm{8}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} }{\mathrm{z}−\mathrm{z}_{\mathrm{k}} }\:\Rightarrow \\ $$$$\int\:\mathrm{F}\left(\mathrm{z}\right)\mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{8}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{7}} \:\mathrm{z}_{\mathrm{k}} ^{\mathrm{2}} \mathrm{ln}\left(\mathrm{z}−\mathrm{z}_{\mathrm{k}} \right)\:+\mathrm{C} \\ $$