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Question Number 37602 by prof Abdo imad last updated on 15/Jun/18
find the value of f(a)= ∫_0 ^∞ ((x^2 −1)/((x^4 +a^4 )^2 ))dx 2) calculate ∫_0 ^∞ ((x^2 −1)/((x^4 +1)^2 ))dx
$${find}\:{the}\:{value}\:{of} \\ $$$${f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:\:\:+{a}^{\mathrm{4}} \right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{{x}^{\mathrm{2}} \:−\mathrm{1}}{\left({x}^{\mathrm{4}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
1) we have f(a)=∫_(−∞) ^(+∞) ((x^2 −1)/((x^4 +a^4 )^2 ))dx let ϕ(z)= ((z^2 −1)/((z^4 +a^4 )^2 )) ϕ(z)= ((z^2 −1)/({ (z^2 )^2 −(ia^2 )^2 }^2 )) =((z^2 −1)/((z^2 −ia^2 )^2 (z^2 +ia^2 )^2 )) =((z^2 −1)/((z −a e^(i(π/4)) )^2 ( z+ae^((iπ)/4) )^2 (z−a e^(−((iπ)/4)) )^2 (z+a e^(−((iπ)/4)) )^2 )) the poles of ϕ are +^− e^(i(π/4)) and +^− e^(−((iπ)/4)) (doubles) let suppose a>0 ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ Res(ϕ, ae^((iπ)/4) ) +Res(ϕ,−ae^(−((iπ)/4)) )} Res(ϕ,ae^(i(π/4)) ) =lim_(z→ae^((iπ)/4) ) (1/((2−1)!)){(z−ae^((iπ)/4) )^2 ϕ(z)}^((1)) =lim_(z→ ae^((iπ)/4) ) { ((z^2 −1)/((z+a e^((iπ)/4) )^2 (z^2 +ia^2 )^2 ))}^((1)) =lim_(z→ ae^((iπ)/4) ) ((2z(z+ae^((iπ)/4) )(z^2 +ia^2 )^2 −(z^2 −1)(2(z+ae^((iπ)/4) )(z^2 +ia^2 )^2 +4z(z^2 +ia^2 )(z+ae^((iπ)/4) )^2 )/((z +ae^((iπ)/4) )^4 ( z^2 +ia^2 )^4 )) =lim_(z→a e^((iπ)/4) ) ((2z(z^2 +ia^2 )−(z^2 −1){ 2(z^2 +ia^2 )+4z(z+ae^((iπ)/4) )})/((z +a e^((iπ)/4) )^3 (z^2 +ia^2 )^3 )) =((2a e^((iπ)/4) (2ia^2 ) −(ia^2 −1){4ia^2 +4ae^((iπ)/4) (2ae^((iπ)/4) )})/((2ae^(i(π/4)) )^3 (2ia^2 )^3 )) =((4ia^3 e^((iπ)/4) −(ia^2 −1){ 4ia^2 +8a^2 i))/(8a^3 i e^((iπ)/4) (−8ia^6 ))) =(( 4ia^3 e^((iπ)/4) −12ia^2 (ia^2 −1))/(64 a^9 e^((iπ)/4) )) =((4ia^(3 ) e^((iπ)/4) +12a^4 +12ia^2 )/(64a^9 e^((iπ)/4) )) =.....
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({a}\right)=\int_{−\infty} ^{+\infty} \:\:\frac{{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)^{\mathrm{2}} }{dx} \\ $$$${let}\:\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} \:−\mathrm{1}}{\left({z}^{\mathrm{4}} \:+{a}^{\mathrm{4}} \right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)=\:\frac{{z}^{\mathrm{2}} \:−\mathrm{1}}{\left\{\:\left({z}^{\mathrm{2}} \right)^{\mathrm{2}} −\left({ia}^{\mathrm{2}} \right)^{\mathrm{2}} \right\}^{\mathrm{2}} } \\ $$$$=\frac{{z}^{\mathrm{2}} \:−\mathrm{1}}{\left({z}^{\mathrm{2}} −{ia}^{\mathrm{2}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$=\frac{{z}^{\mathrm{2}} \:−\mathrm{1}}{\left({z}\:−{a}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left(\:{z}+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}−{a}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}+{a}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} } \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\:\overset{−} {+}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:{and}\:\overset{−} {+}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:\:\left({doubles}\right) \\ $$$${let}\:{suppose}\:{a}>\mathrm{0} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:{Res}\left(\varphi,\:{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−{ae}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\right\} \\ $$$${Res}\left(\varphi,{ae}^{{i}\frac{\pi}{\mathrm{4}}} \right)\:={lim}_{{z}\rightarrow{ae}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\left({z}−{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\:{ae}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\:\left\{\:\:\frac{{z}^{\mathrm{2}} −\mathrm{1}}{\left({z}+{a}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow\:{ae}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\:\frac{\mathrm{2}{z}\left({z}+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{2}} \:−\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}\left({z}+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\mathrm{4z}\left(\mathrm{z}^{\mathrm{2}} \:+\mathrm{ia}^{\mathrm{2}} \right)\left(\mathrm{z}+\mathrm{ae}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \right)^{\mathrm{2}} \right.}{\left({z}\:+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{4}} \left(\:{z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{4}} } \\ $$$$={lim}_{{z}\rightarrow{a}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} } \:\:\frac{\mathrm{2}{z}\left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)−\left({z}^{\mathrm{2}} −\mathrm{1}\right)\left\{\:\mathrm{2}\left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)+\mathrm{4}{z}\left({z}+{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\}}{\left({z}\:+{a}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left({z}^{\mathrm{2}} \:+{ia}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{2}{a}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \left(\mathrm{2}{ia}^{\mathrm{2}} \right)\:−\left({ia}^{\mathrm{2}} −\mathrm{1}\right)\left\{\mathrm{4}{ia}^{\mathrm{2}} \:+\mathrm{4}{ae}^{\frac{{i}\pi}{\mathrm{4}}} \:\left(\mathrm{2}{ae}^{\frac{{i}\pi}{\mathrm{4}}} \right)\right\}}{\left(\mathrm{2}{ae}^{{i}\frac{\pi}{\mathrm{4}}} \right)^{\mathrm{3}} \left(\mathrm{2}{ia}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$$$=\frac{\mathrm{4}{ia}^{\mathrm{3}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:−\left({ia}^{\mathrm{2}} −\mathrm{1}\right)\left\{\:\mathrm{4}{ia}^{\mathrm{2}} \:+\mathrm{8}{a}^{\mathrm{2}} {i}\right)}{\mathrm{8}{a}^{\mathrm{3}} \:{i}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\left(−\mathrm{8}{ia}^{\mathrm{6}} \right)} \\ $$$$=\frac{\:\mathrm{4}{ia}^{\mathrm{3}} {e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:−\mathrm{12}{ia}^{\mathrm{2}} \left({ia}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{64}\:{a}^{\mathrm{9}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} } \\ $$$$=\frac{\mathrm{4}{ia}^{\mathrm{3}\:} {e}^{\frac{{i}\pi}{\mathrm{4}}} \:+\mathrm{12}{a}^{\mathrm{4}} \:\:+\mathrm{12}{ia}^{\mathrm{2}} }{\mathrm{64}{a}^{\mathrm{9}} \:{e}^{\frac{{i}\pi}{\mathrm{4}}} }\:=….. \\ $$
Commented by behi83417@gmail.com last updated on 16/Jun/18
x=a(√(tgt))⇒dx=((a(1+tg^2 t)dt)/(2(√(tgt)))) I=∫((a^2 tgt−1)/((a^4 tg^2 t+a^4 )^2 )).((a(1+tg^2 t)dt)/(2(√(tgt)))) ⇒I=∫((a^2 tgt−1)/(a^8 (tg^2 t+1)^2 )).((a(1+tg^2 t)dt)/(2(√(tgt))))= =(1/a^7 )∫((a^2 tgt−1)/(2(√(tgt))(1+tg^2 t)))dt,tgt=u^2 (1+tg^2 t)dt=2udu⇒dt=((2udu)/(1+u^4 )) ⇒I=(1/a^7 )∫((a^2 u^2 −1)/(2u(1+u^4 ))).((2udu)/(1+u^4 ))=(1/a^7 )∫((a^2 u^2 −1)/((1+u^4 )^2 ))du= =(1/a^7 )[∫((a^2 u^2 )/((1+u^4 )^2 ))du−∫(du/((1+u^4 )^2 ))]=... part#1)∫((u^2 du)/((1+u^4 )^2 ))=(u^3 /(4(u^4 +1)))+((√2)/(16))tg^(−1) ((u(√2))/(1−u^2 ))+ +((√2)/(16))ln((u^2 −u(√2)+1)/( (√(1+u^4 ))))+const part#2)∫(du/((1+u^4 )^2 ))=(u/(4(u^4 +1)))+((3(√2))/(16))tg^(−1) ((u(√2))/(1−u^2 ))− −((3(√2))/(16))ln((u^2 −u(√2)+1)/( (√(1+u^4 ))))+const by symplifing: I=(1/a^5 ).part(#1)−(1/a^7 ).part(#2) change⇒u→(x/a) I=(1/a^5 )[((ax^3 )/(4(a^4 +x^4 )))+((√2)/(16))tg^(−1) (((ax(√2))/(a^2 −x^2 )))− −((3(√2))/(16))ln((x^2 −ax(√2)+a^2 )/( (√(a^4 +x^4 ))))]−(1/a^7 )[((a^3 x)/(4(a^4 +x^4 )))+ +((3(√2))/(16))tg^(−1) ((ax(√2))/(a^2 −x^2 ))−((3(√2))/(16))ln((x^2 −ax(√2)+a^2 )/( (√(a^4 +x^4 ))))]+const 3)I=F(∞)−F(0)=0 4)a=1⇒I=(1/8)[((2x)/(1+x^4 ))+2(√2)tg^(−1) ((x(√2))/(1−x^2 ))− −(√2)ln((x^2 −x(√2)+1)/( (√(1+x^4 ))))]+const .■
$${x}={a}\sqrt{{tgt}}\Rightarrow{dx}=\frac{{a}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{\mathrm{2}\sqrt{{tgt}}} \\ $$$${I}=\int\frac{{a}^{\mathrm{2}} {tgt}−\mathrm{1}}{\left({a}^{\mathrm{4}} {tg}^{\mathrm{2}} {t}+{a}^{\mathrm{4}} \right)^{\mathrm{2}} }.\frac{{a}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{\mathrm{2}\sqrt{{tgt}}} \\ $$$$\Rightarrow{I}=\int\frac{{a}^{\mathrm{2}} {tgt}−\mathrm{1}}{{a}^{\mathrm{8}} \left({tg}^{\mathrm{2}} {t}+\mathrm{1}\right)^{\mathrm{2}} }.\frac{{a}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}}{\mathrm{2}\sqrt{{tgt}}}= \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{7}} }\int\frac{{a}^{\mathrm{2}} {tgt}−\mathrm{1}}{\mathrm{2}\sqrt{{tgt}}\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right)}{dt},{tgt}={u}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+{tg}^{\mathrm{2}} {t}\right){dt}=\mathrm{2}{udu}\Rightarrow{dt}=\frac{\mathrm{2}{udu}}{\mathrm{1}+{u}^{\mathrm{4}} } \\ $$$$ \\ $$$$\Rightarrow{I}=\frac{\mathrm{1}}{{a}^{\mathrm{7}} }\int\frac{{a}^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}{u}\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}.\frac{\mathrm{2}{udu}}{\mathrm{1}+{u}^{\mathrm{4}} }=\frac{\mathrm{1}}{{a}^{\mathrm{7}} }\int\frac{{a}^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} }{du}= \\ $$$$=\frac{\mathrm{1}}{{a}^{\mathrm{7}} }\left[\int\frac{{a}^{\mathrm{2}} {u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} }{du}−\int\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} }\right]=… \\ $$$$\left.\boldsymbol{{part}}#\mathrm{1}\right)\int\frac{{u}^{\mathrm{2}} {du}}{\left(\mathrm{1}+{u}^{\mathrm{4}} \right)^{\mathrm{2}} }=\frac{{u}^{\mathrm{3}} }{\mathrm{4}\left({u}^{\mathrm{4}} +\mathrm{1}\right)}+\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{tg}}^{−\mathrm{1}} \frac{\boldsymbol{{u}}\sqrt{\mathrm{2}}}{\mathrm{1}−\boldsymbol{{u}}^{\mathrm{2}} }+ \\ $$$$+\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{ln}}\frac{\boldsymbol{{u}}^{\mathrm{2}} −\boldsymbol{{u}}\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{u}}^{\mathrm{4}} }}+\boldsymbol{{const}} \\ $$$$\left.\boldsymbol{{part}}#\mathrm{2}\right)\int\frac{\boldsymbol{{du}}}{\left(\mathrm{1}+\boldsymbol{{u}}^{\mathrm{4}} \right)^{\mathrm{2}} }=\frac{\boldsymbol{{u}}}{\mathrm{4}\left(\boldsymbol{{u}}^{\mathrm{4}} +\mathrm{1}\right)}+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{tg}}^{−\mathrm{1}} \frac{\boldsymbol{{u}}\sqrt{\mathrm{2}}}{\mathrm{1}−\boldsymbol{{u}}^{\mathrm{2}} }− \\ $$$$−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{ln}}\frac{\boldsymbol{{u}}^{\mathrm{2}} −\boldsymbol{{u}}\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{u}}^{\mathrm{4}} }}+\boldsymbol{{const}} \\ $$$$\boldsymbol{{by}}\:\boldsymbol{{symplifing}}: \\ $$$$\boldsymbol{{I}}=\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{5}} }.\boldsymbol{{part}}\left(#\mathrm{1}\right)−\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{7}} }.\boldsymbol{{part}}\left(#\mathrm{2}\right) \\ $$$${change}\Rightarrow\boldsymbol{{u}}\rightarrow\frac{\boldsymbol{{x}}}{\boldsymbol{{a}}} \\ $$$${I}=\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{5}} }\left[\frac{\boldsymbol{{ax}}^{\mathrm{3}} }{\mathrm{4}\left(\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{4}} \right)}+\frac{\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{tg}}^{−\mathrm{1}} \left(\frac{\boldsymbol{{ax}}\sqrt{\mathrm{2}}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }\right)−\right. \\ $$$$\left.−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{ln}}\frac{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{ax}}\sqrt{\mathrm{2}}+\boldsymbol{{a}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{4}} }}\right]−\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{7}} }\left[\frac{\boldsymbol{{a}}^{\mathrm{3}} \boldsymbol{{x}}}{\mathrm{4}\left(\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{4}} \right)}+\:\right. \\ $$$$\left.+\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{tg}}^{−\mathrm{1}} \frac{\boldsymbol{{ax}}\sqrt{\mathrm{2}}}{\boldsymbol{{a}}^{\mathrm{2}} −\boldsymbol{{x}}^{\mathrm{2}} }−\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{16}}\boldsymbol{{ln}}\frac{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{ax}}\sqrt{\mathrm{2}}+\boldsymbol{{a}}^{\mathrm{2}} }{\:\sqrt{\boldsymbol{{a}}^{\mathrm{4}} +\boldsymbol{{x}}^{\mathrm{4}} }}\right]+\boldsymbol{{const}} \\ $$$$\left.\mathrm{3}\right)\boldsymbol{{I}}=\boldsymbol{{F}}\left(\infty\right)−\boldsymbol{{F}}\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{4}\right)\boldsymbol{{a}}=\mathrm{1}\Rightarrow\boldsymbol{{I}}=\frac{\mathrm{1}}{\mathrm{8}}\left[\frac{\mathrm{2}\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} }+\mathrm{2}\sqrt{\mathrm{2}}\boldsymbol{{tg}}^{−\mathrm{1}} \frac{\boldsymbol{{x}}\sqrt{\mathrm{2}}}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} }−\right. \\ $$$$\left.−\sqrt{\mathrm{2}}\boldsymbol{{ln}}\frac{\boldsymbol{{x}}^{\mathrm{2}} −\boldsymbol{{x}}\sqrt{\mathrm{2}}+\mathrm{1}}{\:\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{4}} }}\right]+\boldsymbol{{const}}\:.\blacksquare \\ $$
Commented by behi83417@gmail.com last updated on 16/Jun/18
dear prof. abdo! is it ok?
$${dear}\:{prof}.\:{abdo}!\:{is}\:{it}\:{ok}? \\ $$
Commented by prof Abdo imad last updated on 16/Jun/18
sir Behi you have using a good changement at the begining and its seems that your method is correct and i will giveyou the final answer after finishing the calculus...
$${sir}\:{Behi}\:{you}\:{have}\:{using}\:{a}\:{good}\:{changement} \\ $$$${at}\:{the}\:{begining}\:{and}\:{its}\:{seems}\:{that}\:{your}\:{method} \\ $$$${is}\:{correct}\:\:{and}\:{i}\:{will}\:{giveyou}\:{the}\:{final}\:{answer} \\ $$$${after}\:{finishing}\:{the}\:{calculus}… \\ $$

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