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Question Number 60691 by maxmathsup by imad last updated on 24/May/19
calculate f(a) = ∫ (1−(a/x^2 )) arctan(x+(a/x))dx with a real .
$${calculate}\:{f}\left({a}\right)\:=\:\int\:\:\:\left(\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }\right)\:{arctan}\left({x}+\frac{{a}}{{x}}\right){dx}\:\:\:{with}\:{a}\:{real}\:. \\ $$
Commented by maxmathsup by imad last updated on 27/May/19
by parts u^′ =1−(a/x^2 ) and v =arctan(x+(a/x)) ⇒ f(a) =(x+(a/x))arctan(x+(a/x)) −∫ (x+(a/x)) ((1−(a/x^2 ))/(1+(x+(a/x))^2 )) dx =(x+(a/x))arctan(x+(a/x)) −∫ (x+(a/x))((x^2 −a)/(x^2 +(x^2 +a^2 )^2 )) dx but ∫ (x+(a/x)) ((x^2 −a)/(x^2 +(x^2 +a^2 )^2 )) =∫ ((x^4 −a^2 )/(x(x^2 +x^4 +2x^2 a^2 +a^4 ))) dx =∫ ((x^4 −a^2 )/(x( x^4 +(2a^2 +1)x^2 +a^4 )))dx let F(x) =((x^4 −a^2 )/(x(x^4 +(2a^2 +1)x^2 +a^4 ))) poles of F? roots of x^4 +(2a^2 +1)x^2 +a^4 =0 ⇒t^2 +(2a^2 +1)t +a^4 =0 (t =x^2 ) Δ =(2a^2 +1)^2 −4a^4 =4a^4 +4a^2 +1 −4a^4 =4a^2 +1 ⇒ t_1 =((−2a^2 −1 +(√(4a^2 +1)))/2) and t_2 =((−2a^2 −1 −(√(4a^2 +1)))/2) F(x) =((x^4 −a^2 )/(x(x^2 −t_1 )(x^2 −t_2 ))) =((x^4 −a^2 )/(x(x^2 +((2a^2 +1−(√(4a^2 +1)))/2))(x^2 +((2a^2 +1+(√(4a^2 +))1)/2)))) the decomposition of F(x) is at form F(x) =(a/x) +((bx+c)/(x^2 +((2a^2 +1−(√(4a^2 +1)))/2))) +((dx +e)/(x^2 +((2a^2 +1 +(√(4a^2 +1)))/2))) a =lim_(x→a) xF(x) =((a^4 −a^2 )/((a^2 −t_1 )(a^2 −t_2 ))) lim_(x→+∞) xF(x)= 1 =a +b +d ⇒b+d =−a ....be continued....
$${by}\:{parts}\:\:{u}^{'} \:=\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }\:\:{and}\:{v}\:={arctan}\left({x}+\frac{{a}}{{x}}\right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\left({x}+\frac{{a}}{{x}}\right){arctan}\left({x}+\frac{{a}}{{x}}\right)\:−\int\:\:\left({x}+\frac{{a}}{{x}}\right)\:\:\frac{\mathrm{1}−\frac{{a}}{{x}^{\mathrm{2}} }}{\mathrm{1}+\left({x}+\frac{{a}}{{x}}\right)^{\mathrm{2}} }\:{dx} \\ $$$$=\left({x}+\frac{{a}}{{x}}\right){arctan}\left({x}+\frac{{a}}{{x}}\right)\:\:−\int\:\left({x}+\frac{{a}}{{x}}\right)\frac{{x}^{\mathrm{2}} −{a}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dx}\:{but} \\ $$$$\int\:\:\left({x}+\frac{{a}}{{x}}\right)\:\frac{{x}^{\mathrm{2}} −{a}}{{x}^{\mathrm{2}} \:+\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\int\:\:\frac{{x}^{\mathrm{4}} −{a}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:+\mathrm{2}{x}^{\mathrm{2}} {a}^{\mathrm{2}} \:+{a}^{\mathrm{4}} \right)}\:{dx} \\ $$$$=\int\:\:\frac{{x}^{\mathrm{4}} \:−{a}^{\mathrm{2}} }{{x}\left(\:{x}^{\mathrm{4}} \:\:+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{4}} \right)}{dx}\:\:{let}\:{F}\left({x}\right)\:=\frac{{x}^{\mathrm{4}} −{a}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{4}} \right)} \\ $$$${poles}\:{of}\:{F}? \\ $$$${roots}\:{of}\:{x}^{\mathrm{4}} \:+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right){x}^{\mathrm{2}} \:+{a}^{\mathrm{4}} =\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} \:+\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right){t}\:+{a}^{\mathrm{4}} \:=\mathrm{0}\:\:\:\:\left({t}\:={x}^{\mathrm{2}} \right) \\ $$$$\Delta\:=\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{4}{a}^{\mathrm{4}} \:=\mathrm{4}{a}^{\mathrm{4}} \:+\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}\:−\mathrm{4}{a}^{\mathrm{4}} \:=\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}\:\Rightarrow \\ $$$${t}_{\mathrm{1}} =\frac{−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}\:+\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}}\:\:{and}\:\:{t}_{\mathrm{2}} =\frac{−\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}\:−\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}} \\ $$$${F}\left({x}\right)\:=\frac{{x}^{\mathrm{4}} −{a}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({x}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)}\:=\frac{{x}^{\mathrm{4}} −{a}^{\mathrm{2}} }{{x}\left({x}^{\mathrm{2}} +\frac{\mathrm{2}{a}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} \:+\frac{\mathrm{2}{a}^{\mathrm{2}} \:+\mathrm{1}+\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+}\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$\:{the}\:{decomposition}\:{of}\:{F}\left({x}\right)\:{is}\:{at}\:{form} \\ $$$${F}\left({x}\right)\:=\frac{{a}}{{x}}\:+\frac{{bx}+{c}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{2}{a}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}}}\:\:+\frac{{dx}\:+{e}}{{x}^{\mathrm{2}} \:+\frac{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\:+\sqrt{\mathrm{4}{a}^{\mathrm{2}} \:+\mathrm{1}}}{\mathrm{2}}} \\ $$$${a}\:={lim}_{{x}\rightarrow{a}} \:{xF}\left({x}\right)\:=\frac{{a}^{\mathrm{4}} \:−{a}^{\mathrm{2}} }{\left({a}^{\mathrm{2}} −{t}_{\mathrm{1}} \right)\left({a}^{\mathrm{2}} −{t}_{\mathrm{2}} \right)} \\ $$$${lim}_{{x}\rightarrow+\infty} \:{xF}\left({x}\right)=\:\mathrm{1}\:={a}\:+{b}\:+{d}\:\Rightarrow{b}+{d}\:=−{a}\:….{be}\:{continued}…. \\ $$

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