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Question Number 109373 by mathdave last updated on 23/Aug/20

Commented by mnjuly1970 last updated on 23/Aug/20

murray spiegel. advanced calculus.

$${murray}\:{spiegel}.\:{advanced} \\ $$$${calculus}. \\ $$

Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

you want it ?

$$\mathrm{you}\:\mathrm{want}\:\mathrm{it}\:? \\ $$

Commented by bobhans last updated on 23/Aug/20

do you have the pdf file?

Commented by bobhans last updated on 23/Aug/20

yes

$${yes}\: \\ $$

Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

give your email address

$$\mathrm{give}\:\mathrm{your}\:\mathrm{email}\:\mathrm{address} \\ $$

Commented by bobhans last updated on 23/Aug/20

boyhonsome70@gmail.com

$${boyhonsome}\mathrm{70}@{gmail}.{com} \\ $$

Commented by PRITHWISH SEN 2 last updated on 23/Aug/20

please check your mail and reply if you get it.

$$\mathrm{please}\:\mathrm{check}\:\mathrm{your}\:\mathrm{mail}\:\mathrm{and}\:\mathrm{reply}\:\mathrm{if}\:\mathrm{you}\:\mathrm{get}\:\mathrm{it}. \\ $$

Commented by bobhans last updated on 23/Aug/20

yes. thank you very much

$${yes}.\:{thank}\:{you}\:{very}\:{much}\: \\ $$

Commented by mathdave last updated on 23/Aug/20

pls i really need thad advance calculus pdf help me send it through this my email

$${pls}\:{i}\:{really}\:{need}\:{thad}\:{advance}\:{calculus}\:{pdf}\:{help} \\ $$$${me}\:{send}\:{it}\:{through}\:{this}\:{my}\:{email}\: \\ $$

Commented by mathdave last updated on 23/Aug/20

mondaymathew54atgmail.com

$${mondaymathew}\mathrm{54}{atgmail}.{com} \\ $$

Commented by PRITHWISH SEN 2 last updated on 26/Aug/20

something is wrong . I can′t send you. Is the mail address given correct ?

$$\mathrm{something}\:\mathrm{is}\:\mathrm{wrong}\:.\:\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{send}\:\mathrm{you}.\:\mathrm{Is}\:\mathrm{the}\: \\ $$$$\mathrm{mail}\:\mathrm{address}\:\mathrm{given}\:\mathrm{correct}\:? \\ $$

Answered by 1549442205PVT last updated on 23/Aug/20

we find resider of the function f(z)=(z^2 /((z^2 −z+1)^3 )). For this,we need to solve ( z^2 −z+1)^3 =0⇔z^2 −z+1=0 Δ=1−4=3i^2 ⇒z=((1±3i)/2)=e^(±((πi)/3)) Thus,a_1 =e^((πi)/3) and a_2 =e^((−πi)/3) are the poles of degree 3 of f(z).Next,we calculate the residers of f(z) at this poles Res(f(z),e^((πi)/3) )=(1/(2!))lim_(z→e^((πi)/3) ) (d^2 /dz^2 ){(z−e^((πi)/3) ).(z^2 /((z−e^((πi)/3) )^3 (z−e^(−((πi)/3)) )^3 ))} =lim_(z→a_1 ) {(d^2 /dz^2 )[(z^2 /((z−e^((−πi)/3) )^3 ))]}= we have (d/dz)[(z^2 /((z−a)^3 ))]=((2z.(z−a)^3 −3z^2 (z−a)^2 )/z^6 ) ((2z(z−a)−3z^2 )/((z−a)^4 ))=((−z^2 −2az)/((z−a)^4 )).Hence, (d^2 /z^2 )[(z^2 /((z−e^((−πi)/3) )^3 ))]=(((−2z−2a)(z−a)^4 +4(z−a)^3 (z^2 +2az))/((z−a)^8 )) =((−2(z+a)(z−a)+4z^2 +8az)/((z−a)^5 ))=((2z^2 +2a^2 +8az)/((z−a)^5 )) =⇒Res(f(z),e^((πi)/3) )=(1/2).((2.e^((2πi)/3) +2.e^((−2πi)/3) +8e^((πi)/3) .e^((−πi)/3) )/((e^((πi)/3) −e^((−πi)/3) )^5 )) =(1/2).((−1+(−1)+8)/((i(√3))^5 ))=(3/(9i(√3)))=((−i)/(3(√3))) Similarly, Res(f(z),e^((−πi)/3) )=(1/2)lim_(z→e^((−πi)/3) ) {(d^2 /dz^2 )[(z−e^(−((πi)/3)) ).(z^2 /((z−e^((πi)/3) )^3 (z−e^(−((πi)/3)) )^3 ))]} =(1/2).((2.e^((−2πi)/3) +2.e^((2πi)/3) +8e^((−πi)/3) .e^((πi)/3) )/((e^((−πi)/3) −e^((πi)/3) )^5 ))= =(1/2).(((−1)+(−1)+8)/((−i(√3))^5 ))=(3/(9i(√3)))=((−i)/(3(√3))) Σ_a_k Res(f(z))=((−2i)/( (√3))).Therefore, ∫_(−∞) ^(+∞) f(x)dx=2πiΣRes(f(z),a_k ) =2πi×((−2i)/(3(√3)))=((4π)/(3(√3)))

$$\mathrm{we}\:\mathrm{find}\:\mathrm{resider}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}\: \\ $$$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\right)^{\mathrm{3}} }.\:\mathrm{For}\:\mathrm{this},\mathrm{we}\:\mathrm{need}\:\mathrm{to} \\ $$$$\mathrm{solve}\:\:\:\:\left(\:\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0}\Leftrightarrow\mathrm{z}^{\mathrm{2}} −\mathrm{z}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=\mathrm{3i}^{\mathrm{2}} \Rightarrow\mathrm{z}=\frac{\mathrm{1}\pm\mathrm{3i}}{\mathrm{2}}=\mathrm{e}^{\pm\frac{\pi\mathrm{i}}{\mathrm{3}}} \\ $$$$\mathrm{Thus},\mathrm{a}_{\mathrm{1}} =\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \mathrm{and}\:\mathrm{a}_{\mathrm{2}} =\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} \mathrm{are}\:\mathrm{the}\:\mathrm{poles} \\ $$$$\mathrm{of}\:\mathrm{degree}\:\mathrm{3}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{z}\right).\mathrm{Next},\mathrm{we}\:\mathrm{calculate}\:\mathrm{the}\:\mathrm{residers}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{z}\right)\:\mathrm{at}\:\mathrm{this}\:\mathrm{poles} \\ $$$$\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right),\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)=\frac{\mathrm{1}}{\mathrm{2}!}\underset{\mathrm{z}\rightarrow\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} } {\mathrm{lim}}\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }\left\{\left(\mathrm{z}−\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right).\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} }\right\} \\ $$$$=\underset{\mathrm{z}\rightarrow\mathrm{a}_{\mathrm{1}} } {\mathrm{lim}}\left\{\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }\left[\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} }\right]\right\}= \\ $$$$\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{d}}{\mathrm{dz}}\left[\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{3}} }\right]=\frac{\mathrm{2z}.\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{3}} −\mathrm{3z}^{\mathrm{2}} \left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{z}^{\mathrm{6}} } \\ $$$$\frac{\mathrm{2z}\left(\mathrm{z}−\mathrm{a}\right)−\mathrm{3z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{4}} }=\frac{−\mathrm{z}^{\mathrm{2}} −\mathrm{2az}}{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{4}} }.\mathrm{Hence}, \\ $$$$\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{z}^{\mathrm{2}} }\left[\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} }\right]=\frac{\left(−\mathrm{2z}−\mathrm{2a}\right)\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{4}} +\mathrm{4}\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{3}} \left(\mathrm{z}^{\mathrm{2}} +\mathrm{2az}\right)}{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{8}} } \\ $$$$=\frac{−\mathrm{2}\left(\mathrm{z}+\mathrm{a}\right)\left(\mathrm{z}−\mathrm{a}\right)+\mathrm{4z}^{\mathrm{2}} +\mathrm{8az}}{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{5}} }=\frac{\mathrm{2z}^{\mathrm{2}} +\mathrm{2a}^{\mathrm{2}} +\mathrm{8az}}{\left(\mathrm{z}−\mathrm{a}\right)^{\mathrm{5}} } \\ $$$$=\Rightarrow\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right),\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}.\mathrm{e}^{\frac{\mathrm{2}\pi\mathrm{i}}{\mathrm{3}}} +\mathrm{2}.\mathrm{e}^{\frac{−\mathrm{2}\pi\mathrm{i}}{\mathrm{3}}} +\mathrm{8e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} .\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} }{\left(\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} −\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{5}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{−\mathrm{1}+\left(−\mathrm{1}\right)+\mathrm{8}}{\left(\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }=\frac{\mathrm{3}}{\mathrm{9i}\sqrt{\mathrm{3}}}=\frac{−\mathrm{i}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{Similarly}, \\ $$$$\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right),\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} \right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{z}\rightarrow\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} } {\mathrm{lim}}\left\{\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dz}^{\mathrm{2}} }\left[\left(\mathrm{z}−\mathrm{e}^{−\frac{\pi\mathrm{i}}{\mathrm{3}}} \right).\frac{\mathrm{z}^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} \left(\mathrm{z}−\mathrm{e}^{−\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{3}} }\right]\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\mathrm{2}.\mathrm{e}^{\frac{−\mathrm{2}\pi\mathrm{i}}{\mathrm{3}}} +\mathrm{2}.\mathrm{e}^{\frac{\mathrm{2}\pi\mathrm{i}}{\mathrm{3}}} +\mathrm{8e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} .\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} }{\left(\mathrm{e}^{\frac{−\pi\mathrm{i}}{\mathrm{3}}} −\mathrm{e}^{\frac{\pi\mathrm{i}}{\mathrm{3}}} \right)^{\mathrm{5}} }= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\left(−\mathrm{1}\right)+\left(−\mathrm{1}\right)+\mathrm{8}}{\left(−\mathrm{i}\sqrt{\mathrm{3}}\right)^{\mathrm{5}} }=\frac{\mathrm{3}}{\mathrm{9i}\sqrt{\mathrm{3}}}=\frac{−\mathrm{i}}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\underset{\mathrm{a}_{\mathrm{k}} } {\sum}\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right)\right)=\frac{−\mathrm{2i}}{\:\sqrt{\mathrm{3}}}.\mathrm{Therefore}, \\ $$$$\int_{−\infty} ^{+\infty} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{2}\pi\mathrm{i}\Sigma\mathrm{Res}\left(\mathrm{f}\left(\mathrm{z}\right),\mathrm{a}_{\mathrm{k}} \right) \\ $$$$=\mathrm{2}\pi\mathrm{i}×\frac{−\mathrm{2i}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{4}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

Commented by mathdave last updated on 23/Aug/20

recheck ur wirking well

$${recheck}\:{ur}\:{wirking}\:{well}\: \\ $$

Commented by 1549442205PVT last updated on 24/Aug/20

Thsnk you,sir.You are welcome

$$\mathrm{Thsnk}\:\mathrm{you},\mathrm{sir}.\mathrm{You}\:\mathrm{are}\:\mathrm{welcome} \\ $$

Answered by mathmax by abdo last updated on 23/Aug/20

I =∫_(−∞) ^(+∞) (x^2 /((x^2 −x+1)^3 ))dx we hsve x^2 −x +1 =(x−(1/2))^2 +(3/4) we do the changement x−(1/2) =((√3)/2)t ⇒ I =∫_(−∞) ^(+∞) ((1(1+(√3)t)^2 )/(4((3/4))^2 (t^2 +1)^3 )) ×((√3)/2) dt =((2(√3))/9)∫_(−∞) ^(+∞) (((1+(√3)z)^2 )/((t^2 +1)^3 )) dt let ϕ(z) =(((1+(√3)z)^2 )/((z^2 +1)^3 )) ⇒ ϕ(z) =(((1+(√3)z)^2 )/((z−i)^3 (z+i)^3 )) residus theorem ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) =lim_(z→i) (1/((3−1)!)){(z−i)^3 ϕ(z)}^((2)) =(1/2)lim_(z→i) {(((1+(√3)z)^2 )/((z+i)^3 ))}^((2)) =(1/2)lim_(z→i) { ((2(√3)(1+(√3)z)(z+i)^3 −3(z+i)^2 (1+(√3)z)^2 )/((z+i)^6 ))}^((1)) =(1/2)lim_(z→i) {((2(√3)(1+(√3)z)(z+i)−3(1+(√3)z)^2 )/((z+i)^4 ))}^((1)) =(1/2)lim_(z→i) (((6(z+i)+2(√3)(1+(√3)z)−6(√3)(1+(√3)z))(z+i)^4 −4(z+i)^3 (2(√3)(1+(√3)z)(z+i)−3(1+(√3)z)^2 )/((z+i)^8 )) rest to finish the calculus....

$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{x}^{\mathrm{2}} }{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)^{\mathrm{3}} }\mathrm{dx}\:\:\mathrm{we}\:\mathrm{hsve}\:\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}\:+\mathrm{1}\:=\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\int_{−\infty} ^{+\infty} \:\frac{\mathrm{1}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{t}\right)^{\mathrm{2}} }{\mathrm{4}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} \left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\int_{−\infty} ^{+\infty} \:\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{t}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\mathrm{dt}\:\:\mathrm{let}\:\varphi\left(\mathrm{z}\right)\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} }\:\Rightarrow \\ $$$$\varphi\left(\mathrm{z}\right)\:=\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\:\mathrm{residus}\:\mathrm{theorem}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left(\mathrm{z}\right)\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\varphi,\mathrm{i}\right) \\ $$$$\mathrm{Res}\left(\varphi,\mathrm{i}\right)\:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−\mathrm{1}\right)!}\left\{\left(\mathrm{z}−\mathrm{i}\right)^{\mathrm{3}} \varphi\left(\mathrm{z}\right)\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\frac{\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} }\right\}^{\left(\mathrm{2}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\:\left\{\:\frac{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{2}} \left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{6}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\left\{\frac{\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} }{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{i}} \:\:\:\frac{\left(\mathrm{6}\left(\mathrm{z}+\mathrm{i}\right)+\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)−\mathrm{6}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)\right)\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{4}} −\mathrm{4}\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{3}} \left(\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)\left(\mathrm{z}+\mathrm{i}\right)−\mathrm{3}\left(\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{z}\right)^{\mathrm{2}} \right.}{\left(\mathrm{z}+\mathrm{i}\right)^{\mathrm{8}} } \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{finish}\:\mathrm{the}\:\mathrm{calculus}.... \\ $$

Answered by Sarah85 last updated on 24/Aug/20

remember Ostrogradski′s Method introduced by former member MJS ⇒ ∫(x^2 /((x^2 −x+1)^3 ))dx=((2x^3 −3x^2 +2x−2)/((x^2 −x+1)^2 ))+(1/3)∫(dx/(x^2 −x+1))= =((2x^3 −3x^2 +2x−2)/((x^2 −x+1)^2 ))+((2(√3))/9)arctan (((2x−1)(√3))/( 3)) +C ⇒ ∫_(−∞) ^(+∞) (x^2 /((x^2 −x+1)^3 ))dx=[((2(√3))/9)arctan (((2x−1)(√3))/( 3))]_(−∞) ^(+∞) = =((2π(√3))/9)

$$\mathrm{remember}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{Method}\:\mathrm{introduced} \\ $$$$\mathrm{by}\:\mathrm{former}\:\mathrm{member}\:\mathrm{MJS} \\ $$$$\Rightarrow \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}=\frac{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}= \\ $$$$=\frac{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{2}}{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\:\mathrm{3}}\:+{C} \\ $$$$\Rightarrow \\ $$$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}=\left[\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\mathrm{arctan}\:\frac{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{\mathrm{3}}}{\:\mathrm{3}}\right]_{−\infty} ^{+\infty} = \\ $$$$=\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}} \\ $$

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