Proof ∫(1/(a^2 −x^2 ))dx =(1/(2a))ln∣((a+x)/(a−x))∣+c


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Question Number 27271 by GANGADHARSETHI last updated on 04/Jan/18

$P r o o f$ $\int \frac{1}{a^{2} - x^{2}} d x = \frac{1}{2 a} l n ∣ \frac{a + x}{a - x} ∣ + c$
	Answered by sma3l2996 last updated on 04/Jan/18

	$\int \frac{d x}{a^{2} - x^{2}} = \int \frac{d x}{(a - x) (a + x)}$ $\frac{1}{(x + a) (a - x)} = \frac{α}{a + x} + \frac{β}{a - x}; α = \frac{1}{2 a} = β$ $s o : \int \frac{d x}{a^{2} - x^{2}} = \frac{1}{2 a} \int (\frac{1}{a + x} + \frac{1}{a - x}) d x$ $= \frac{1}{2 a} (l n ∣ a + x ∣ - l n ∣ a - x ∣) + c$ $= \frac{1}{2 a} l n ∣ \frac{a + x}{a - x} ∣ + c$
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