Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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.

$${nvermind}\:{sir}\:{joel}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com