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Question Number 83085 by mathmax by abdo last updated on 27/Feb/20
1) find ∫ (dx/((x^2 +1)^4 )) 2)calculate ∫_0 ^∞ (dx/((x^2 +1)^4 ))
$$\left.\mathrm{1}\right)\:{find}\:\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} } \\ $$
Commented by abdomathmax last updated on 28/Feb/20
2) let I=∫_0 ^∞ (dx/((x^2 +1)^4 ))dx ⇒2I=∫_(−∞) ^(+∞) (dx/((x^2 +1)^4 )) let W(z)=(1/((z^2 +1)^4 )) ⇒W(z)=(1/((z−i)^4 (z+i)^4 )) ∫_(−∞) ^(+∞) W(z)dz =2iπ Res(W,i) Res(W,i)=lim_(z→i) (1/((4−1)!)){(z−i)^4 W(z)}^((3)) =lim_(z→i) (1/6){(z+i)^(−4) }^((3)) =lim_(z→i) ((−2)/3){(z+i)^(−5) }^((2)) =lim_(z→i) ((10)/3){(z+i)^(−6) }^()1)) =lim_(z→i) (−20)(z+i)^(−7) =((−20)/((2i)^7 )) =((−20)/(2^7 ×i^(8−1) )) =((−20i)/2^7 ) =((−20i)/(2^5 ×4)) =((−20i)/(32×4)) =−((4×5i)/(32×4)) =−((5i)/(32)) ⇒∫_(−∞) ^(+∞) W(z)dz =2iπ×(((−5i))/(32))) =((5π)/(16)) =2I ⇒ I =((5π)/(32))
$$\left.\mathrm{2}\right)\:{let}\:{I}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} }{dx}\:\Rightarrow\mathrm{2}{I}=\int_{−\infty} ^{+\infty} \:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} } \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{i}\right)^{\mathrm{4}} \left({z}+{i}\right)^{\mathrm{4}} } \\ $$$$\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{i}\right) \\ $$$${Res}\left({W},{i}\right)={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{1}}{\left(\mathrm{4}−\mathrm{1}\right)!}\left\{\left({z}−{i}\right)^{\mathrm{4}} {W}\left({z}\right)\right\}^{\left(\mathrm{3}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\:\frac{\mathrm{1}}{\mathrm{6}}\left\{\left({z}+{i}\right)^{−\mathrm{4}} \right\}^{\left(\mathrm{3}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{−\mathrm{2}}{\mathrm{3}}\left\{\left({z}+{i}\right)^{−\mathrm{5}} \right\}^{\left(\mathrm{2}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\frac{\mathrm{10}}{\mathrm{3}}\left\{\left({z}+{i}\right)^{−\mathrm{6}} \right\}^{\left.\right)\left.\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}} \:\:\left(−\mathrm{20}\right)\left({z}+{i}\right)^{−\mathrm{7}} \\ $$$$=\frac{−\mathrm{20}}{\left(\mathrm{2}{i}\right)^{\mathrm{7}} }\:=\frac{−\mathrm{20}}{\mathrm{2}^{\mathrm{7}} ×{i}^{\mathrm{8}−\mathrm{1}} }\:=\frac{−\mathrm{20}{i}}{\mathrm{2}^{\mathrm{7}} }\:=\frac{−\mathrm{20}{i}}{\mathrm{2}^{\mathrm{5}} ×\mathrm{4}}\:=\frac{−\mathrm{20}{i}}{\mathrm{32}×\mathrm{4}} \\ $$$$=−\frac{\mathrm{4}×\mathrm{5}{i}}{\mathrm{32}×\mathrm{4}}\:=−\frac{\mathrm{5}{i}}{\mathrm{32}}\:\Rightarrow\int_{−\infty} ^{+\infty} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi×\left(\frac{\left.−\mathrm{5}{i}\right)}{\mathrm{32}}\right) \\ $$$$=\frac{\mathrm{5}\pi}{\mathrm{16}}\:=\mathrm{2}{I}\:\Rightarrow\:{I}\:=\frac{\mathrm{5}\pi}{\mathrm{32}} \\ $$
Commented by mathmax by abdo last updated on 28/Feb/20
1)complex method let A =∫ (dx/((x^2 +1)^4 )) ⇒A =∫ (dx/((x−i)^4 (x+i)^4 )) =∫ (dx/((((x−i)/(x+i)))^4 (x+i)^8 )) let use the changement ((x−i)/(x+i))=t ⇒ x−i=xt+it ⇒(1−t)x=i(1+t) ⇒x=i((1+t)/(1−t)) ⇒dx=i((1−t+(1+t))/((1−t)^2 ))dt =((2i)/((t−1)^2 ))dt also x+i =((i+it)/(1−t))+i =((i+it+i−it)/(1−t)) =((2i)/(1−t)) ⇒ A =∫ ((2idt)/((t−1)^2 ×t^4 ×(((2i)/(t−1)))^8 )) =(1/((2i)^7 ))∫ (((t−1)^8 dt)/((t−1)^2 t^4 )) =(i/2^7 )∫ (((t−1)^6 )/t^4 )dt =(i/2^7 )∫ ((Σ_(k=0) ^6 C_6 ^k (−1)^k t^k )/t^4 )dt =(i/2^7 )∫ Σ_(k=0) ^6 (−1)^k C_6 ^k t^(k−4) dt =(i/2^7 ){Σ_(k=0 andk≠3) ^6 (−1)^k C_6 ^k ×(1/(k−3))t^(k−3) −C_6 ^3 ln(t)}+C ⇒A =(i/2^7 ){Σ_(k=0 and k≠3) ^6 (((−1)^k C_6 ^k )/(k−3)) (((x−i)/(x+i)))^(k−3) −C_6 ^3 ln(((x−i)/(x+i)))} +C
$$\left.\mathrm{1}\right){complex}\:{method}\:{let}\:{A}\:=\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{4}} }\:\Rightarrow{A}\:=\int\:\:\frac{{dx}}{\left({x}−{i}\right)^{\mathrm{4}} \left({x}+{i}\right)^{\mathrm{4}} } \\ $$$$=\int\:\:\:\frac{{dx}}{\left(\frac{{x}−{i}}{{x}+{i}}\right)^{\mathrm{4}} \left({x}+{i}\right)^{\mathrm{8}} }\:{let}\:{use}\:{the}\:{changement}\:\frac{{x}−{i}}{{x}+{i}}={t}\:\Rightarrow \\ $$$${x}−{i}={xt}+{it}\:\Rightarrow\left(\mathrm{1}−{t}\right){x}={i}\left(\mathrm{1}+{t}\right)\:\Rightarrow{x}={i}\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\Rightarrow{dx}={i}\frac{\mathrm{1}−{t}+\left(\mathrm{1}+{t}\right)}{\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{2}{i}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:{also}\:\:{x}+{i}\:=\frac{{i}+{it}}{\mathrm{1}−{t}}+{i}\:=\frac{{i}+{it}+{i}−{it}}{\mathrm{1}−{t}}\:=\frac{\mathrm{2}{i}}{\mathrm{1}−{t}}\:\Rightarrow \\ $$$${A}\:=\int\:\:\:\frac{\mathrm{2}{idt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} ×{t}^{\mathrm{4}} ×\left(\frac{\mathrm{2}{i}}{{t}−\mathrm{1}}\right)^{\mathrm{8}} }\:=\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{7}} }\int\:\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{8}} \:{dt}}{\left({t}−\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{4}} } \\ $$$$=\frac{{i}}{\mathrm{2}^{\mathrm{7}} }\int\:\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{6}} }{{t}^{\mathrm{4}} }{dt}\:=\frac{{i}}{\mathrm{2}^{\mathrm{7}} }\int\:\:\frac{\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \left(−\mathrm{1}\right)^{{k}} \:{t}^{{k}} }{{t}^{\mathrm{4}} }{dt} \\ $$$$=\frac{{i}}{\mathrm{2}^{\mathrm{7}} }\int\:\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{6}} ^{{k}} \:{t}^{{k}−\mathrm{4}} \:{dt} \\ $$$$=\frac{{i}}{\mathrm{2}^{\mathrm{7}} }\left\{\sum_{{k}=\mathrm{0}\:{andk}\neq\mathrm{3}} ^{\mathrm{6}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{6}} ^{{k}} \:×\frac{\mathrm{1}}{{k}−\mathrm{3}}{t}^{{k}−\mathrm{3}} \:−{C}_{\mathrm{6}} ^{\mathrm{3}} \:{ln}\left({t}\right)\right\}+{C} \\ $$$$\Rightarrow{A}\:=\frac{{i}}{\mathrm{2}^{\mathrm{7}} }\left\{\sum_{{k}=\mathrm{0}\:{and}\:{k}\neq\mathrm{3}} ^{\mathrm{6}} \frac{\left(−\mathrm{1}\right)^{{k}} \:{C}_{\mathrm{6}} ^{{k}} }{{k}−\mathrm{3}}\:\left(\frac{{x}−{i}}{{x}+{i}}\right)^{{k}−\mathrm{3}} \:−{C}_{\mathrm{6}} ^{\mathrm{3}} {ln}\left(\frac{{x}−{i}}{{x}+{i}}\right)\right\}\:+{C} \\ $$
Answered by mind is power last updated on 27/Feb/20
∫(dx/((x^2 +a)))=f(a) x=(√a) u⇒∫(((√a)du)/(a(1+u^2 )))=((arctan(u))/( (√a)))+c ∫(dx/(x^2 +a))=((arctan((x/( (√a)))))/( (√a)))+c=f(a) ∫(dx/((x^2 +a)^4 ))=−((f^4 (a))/6)
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}\right)}={f}\left({a}\right) \\ $$$${x}=\sqrt{{a}}\:{u}\Rightarrow\int\frac{\sqrt{{a}}{du}}{{a}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}=\frac{{arctan}\left({u}\right)}{\:\sqrt{{a}}}+{c} \\ $$$$\int\frac{{dx}}{{x}^{\mathrm{2}} +{a}}=\frac{{arctan}\left(\frac{{x}}{\:\sqrt{{a}}}\right)}{\:\sqrt{{a}}}+{c}={f}\left({a}\right) \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{a}\right)^{\mathrm{4}} }=−\frac{{f}^{\mathrm{4}} \left({a}\right)}{\mathrm{6}} \\ $$

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