Menu Close

Question-134052




Question Number 134052 by shaker last updated on 27/Feb/21
Answered by mathmax by abdo last updated on 27/Feb/21
I =∫(x^3 /(x^6  +3))dx  ⇒I =∫  (x^3 /(x^6  +(3^(1/6) )^6 ))dx =_(x=3^(1/6) t)   ∫ ((3^(1/2) t^3 )/(3(1+t^6 ))) 3^(1/6)  dt  =3^((1/2)+(1/6)−1)  ∫  (t^3 /(t^6  +1))dt =3^(−(1/3))  ∫  (t^3 /(t^6  +1)) dt    z^6  +1=0 ⇒z^6  =e^(i(π+2kπ))  =e^(i(2k+1)π)  ⇒z_k =e^((i(2k+1)π)/6)  and  k∈[[0,5]] ⇒(t^3 /(t^6  +1)) =(t^3 /(Π_(k=0) ^5 (t−z_k )))=Σ_(k=0) ^5  (a_k /(t−z_k ))  a_k =(z_k ^3 /(6z_k ^5 )) =(z_k ^4 /(−6))=−(1/6) z_k ^4  ⇒(t^3 /(t^6  +1))=−(1/6)Σ_(k=0) ^5  (e^((i(2k+1)2π)/3) /(t−e^((i(2k+1)π)/6) )) ⇒  ∫  (t^3 /(1+t^6 ))dt =−(1/6)Σ_(k=0) ^5  e^((2πi(2k+1))/3) ln(t−e^((i(2k+1)π)/6) ) +C
$$\mathrm{I}\:=\int\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{6}} \:+\mathrm{3}}\mathrm{dx}\:\:\Rightarrow\mathrm{I}\:=\int\:\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{6}} \:+\left(\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\mathrm{6}} }\mathrm{dx}\:=_{\mathrm{x}=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{6}}} \mathrm{t}} \:\:\int\:\frac{\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{t}^{\mathrm{3}} }{\mathrm{3}\left(\mathrm{1}+\mathrm{t}^{\mathrm{6}} \right)}\:\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{6}}} \:\mathrm{dt} \\ $$$$=\mathrm{3}^{\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:\int\:\:\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{6}} \:+\mathrm{1}}\mathrm{dt}\:=\mathrm{3}^{−\frac{\mathrm{1}}{\mathrm{3}}} \:\int\:\:\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{6}} \:+\mathrm{1}}\:\mathrm{dt}\:\: \\ $$$$\mathrm{z}^{\mathrm{6}} \:+\mathrm{1}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{6}} \:=\mathrm{e}^{\mathrm{i}\left(\pi+\mathrm{2k}\pi\right)} \:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi} \:\Rightarrow\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \:\mathrm{and} \\ $$$$\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{5}\right]\right]\:\Rightarrow\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{6}} \:+\mathrm{1}}\:=\frac{\mathrm{t}^{\mathrm{3}} }{\prod_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \left(\mathrm{t}−\mathrm{z}_{\mathrm{k}} \right)}=\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{a}_{\mathrm{k}} }{\mathrm{t}−\mathrm{z}_{\mathrm{k}} } \\ $$$$\mathrm{a}_{\mathrm{k}} =\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{3}} }{\mathrm{6z}_{\mathrm{k}} ^{\mathrm{5}} }\:=\frac{\mathrm{z}_{\mathrm{k}} ^{\mathrm{4}} }{−\mathrm{6}}=−\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{z}_{\mathrm{k}} ^{\mathrm{4}} \:\Rightarrow\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{t}^{\mathrm{6}} \:+\mathrm{1}}=−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\frac{\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\mathrm{2}\pi}{\mathrm{3}}} }{\mathrm{t}−\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} }\:\Rightarrow \\ $$$$\int\:\:\frac{\mathrm{t}^{\mathrm{3}} }{\mathrm{1}+\mathrm{t}^{\mathrm{6}} }\mathrm{dt}\:=−\frac{\mathrm{1}}{\mathrm{6}}\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{5}} \:\mathrm{e}^{\frac{\mathrm{2}\pi\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)}{\mathrm{3}}} \mathrm{ln}\left(\mathrm{t}−\mathrm{e}^{\frac{\mathrm{i}\left(\mathrm{2k}+\mathrm{1}\right)\pi}{\mathrm{6}}} \right)\:+\mathrm{C} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *