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Question Number 34237 by abdo mathsup 649 cc last updated on 03/May/18
find ∫ (dx/(x^2 −a)) with a ∈ C .
$${find}\:\:\int\:\:\:\frac{{dx}}{{x}^{\mathrm{2}} \:−{a}}\:\:{with}\:{a}\:\in\:{C}\:. \\ $$
Commented by Joel578 last updated on 04/May/18
thank you for correction Sir
$${thank}\:{you}\:{for}\:{correction}\:{Sir} \\ $$
Commented by abdo mathsup 649 cc last updated on 04/May/18
let put (√a)=α +iβ with (α,β) ∈ R^2 I = (1/(2(√a)))∫ ( (1/(x−(√a))) −(1/(x+(√a))))dx = (1/(2(α+iβ))) ( ∫ (dx/(x−(√a))) − ∫ (dx/(x+(√a)))) but ∫ (dx/(x−(√a))) = ∫ (dx/(x−α −iβ)) = ∫ ((x−α +i β)/((x−α)^2 +β^2 ))dx = ∫ ((x−α)/((x−α)^2 +β^2 ))dx +iβ ∫ (dx/((x−α)^2 +β^2 )) but we have ∫ ((x−α)/((x −α)^2 +β^2 ))dx = (1/2)ln( (x−α)^2 +β^2 ) +c_1 changement x−α=βt give ∫ (dx/((x−α)^2 +β^2 )) = ∫ ((βdt)/(β^2 (1+t^2 ))) =(1/β) arctant +c_2 ⇒ ∫ (dx/(x−(√a))) = (1/2)ln((x−α)^2 +β^2 ) +i arctant +λ we folow the same method to find ∫ (dx/(x +(√a))) ...
$${let}\:{put}\:\sqrt{{a}}=\alpha\:+{i}\beta\:\:\:{with}\:\left(\alpha,\beta\right)\:\in\:{R}^{\mathrm{2}} \\ $$$${I}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}\int\:\:\left(\:\frac{\mathrm{1}}{{x}−\sqrt{{a}}}\:−\frac{\mathrm{1}}{{x}+\sqrt{{a}}}\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\left(\alpha+{i}\beta\right)}\:\:\left(\:\int\:\:\:\frac{{dx}}{{x}−\sqrt{{a}}}\:\:−\:\int\:\:\:\frac{{dx}}{{x}+\sqrt{{a}}}\right)\:{but} \\ $$$$\int\:\:\:\frac{{dx}}{{x}−\sqrt{{a}}}\:=\:\int\:\:\:\:\frac{{dx}}{{x}−\alpha\:−{i}\beta} \\ $$$$=\:\int\:\:\:\:\frac{{x}−\alpha\:+{i}\:\beta}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dx}\:=\:\int\:\:\:\:\frac{{x}−\alpha}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dx} \\ $$$$+{i}\beta\:\int\:\:\:\:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:\:{but}\:{we}\:{have} \\ $$$$\int\:\:\:\:\frac{{x}−\alpha}{\left({x}\:−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\:\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+{c}_{\mathrm{1}} \\ $$$${changement}\:\:{x}−\alpha=\beta{t}\:\:{give} \\ $$$$\int\:\:\:\:\:\:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:=\:\int\:\:\:\:\frac{\beta{dt}}{\beta^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)}\:=\frac{\mathrm{1}}{\beta}\:{arctant}\:+{c}_{\mathrm{2}} \\ $$$$\Rightarrow\:\int\:\:\:\:\frac{{dx}}{{x}−\sqrt{{a}}}\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:\:+{i}\:{arctant}\:+\lambda \\ $$$${we}\:{folow}\:{the}\:{same}\:{method}\:{to}\:{find}\:\:\:\int\:\:\:\:\frac{{dx}}{{x}\:+\sqrt{{a}}}\:\:… \\ $$
Commented by math khazana by abdo last updated on 04/May/18
∫ (dx/((x−α)^2 +β^2 )) =(1/β) arctan(((x−α)/β)) +c_2 ⇒ ∫ (dx/(x−(√a))) = (1/2)ln((x−α)^2 +β^2 ) +i arctan(((x−α)/β)) +c
$$\int\:\:\:\:\:\:\frac{{dx}}{\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\beta}\:{arctan}\left(\frac{{x}−\alpha}{\beta}\right)\:+{c}_{\mathrm{2}} \:\:\Rightarrow \\ $$$$\int\:\:\:\:\:\frac{{dx}}{{x}−\sqrt{{a}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\left({x}−\alpha\right)^{\mathrm{2}} \:+\beta^{\mathrm{2}} \right)\:+{i}\:{arctan}\left(\frac{{x}−\alpha}{\beta}\right)\:+{c} \\ $$
Commented by abdo mathsup 649 cc last updated on 04/May/18
nvermind sir joel.
$${nvermind}\:{sir}\:{joel}. \\ $$

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