Mathematics

Integrals of sum particular function 
prove that $$\displaystyle\int {{{dx} \over {{x^2} - {a^2}}} = {1 \over {2a}}\log \left| {{{x - a} \over {x + a}}} \right| + c} $$


SOLUTION
Since $$\dfrac{1}{x^2-a^2}=\dfrac{1}{(x-a)(x+a)}=\dfrac{1}{2a}\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]$$

Now, taking

$$\displaystyle\int \dfrac{dx}{x^2-a^2}=\displaystyle\int \dfrac{1}{2a}\left[\dfrac{1}{x-a}-\dfrac{1}{x+a}\right]dx$$

$$=\dfrac{1}{2a}\left[\displaystyle\int \dfrac{dx}{x-a}-\displaystyle\int \dfrac{dx}{x+a}\right]$$ [using formula $$\displaystyle\int \dfrac{dx}{x}=log x$$]

$$=\dfrac{1}{2a}[log |x-a|-log|x+a|]+c$$

$$=\dfrac{1}{2a}log\left|\dfrac{x-a}{x+a}\right|+c$$ proved.
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Subjective Medium Published on 17th 09, 2020
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