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Question Number 34237 by abdo mathsup 649 cc last updated on 03/May/18

$$find\int \frac{dx}{x^2-a}witha\in C.$$

Commented by Joel578 last updated on 04/May/18

$$thankyouforcorrectionSir$$

Commented by abdo mathsup 649 cc last updated on 04/May/18

$$letput\sqrt{a}=\alpha +i\beta with(\alpha ,\beta )\in R^2$$$$I=\frac{1}{2\sqrt{a}}\int (\frac{1}{x-\sqrt{a}}-\frac{1}{x+\sqrt{a}})dx$$$$=\frac{1}{2(\alpha +i\beta )}(\int \frac{dx}{x-\sqrt{a}}-\int \frac{dx}{x+\sqrt{a}})but$$$$\int \frac{dx}{x-\sqrt{a}}=\int \frac{dx}{x-\alpha -i\beta }$$$$=\int \frac{x-\alpha +i\beta }{{(x-\alpha )}^2+{\beta }^2}dx=\int \frac{x-\alpha }{{(x-\alpha )}^2+{\beta }^2}dx$$$$+i\beta \int \frac{dx}{{(x-\alpha )}^2+{\beta }^2}butwehave$$$$\int \frac{x-\alpha }{{(x-\alpha )}^2+{\beta }^2}dx=\frac{1}{2}ln({(x-\alpha )}^2+{\beta }^2)+c_1$$$$changementx-\alpha =\beta tgive$$$$\int \frac{dx}{{(x-\alpha )}^2+{\beta }^2}=\int \frac{\beta dt}{{\beta }^2(1+t^2)}=\frac{1}{\beta }arctant+c_2$$$$\Rightarrow \int \frac{dx}{x-\sqrt{a}}=\frac{1}{2}ln({(x-\alpha )}^2+{\beta }^2)+iarctant+\lambda$$$$wefolowthesamemethodtofind\int \frac{dx}{x+\sqrt{a}}...$$

Commented by math khazana by abdo last updated on 04/May/18

$$\int \frac{dx}{{(x-\alpha )}^2+{\beta }^2}=\frac{1}{\beta }arctan\left(\frac{x-\alpha }{\beta }\right)+c_2\Rightarrow$$$$\int \frac{dx}{x-\sqrt{a}}=\frac{1}{2}ln({(x-\alpha )}^2+{\beta }^2)+iarctan\left(\frac{x-\alpha }{\beta }\right)+c$$

Commented by abdo mathsup 649 cc last updated on 04/May/18

$$nvermindsirjoel.$$

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