The integral rule for multiplicative inverse of difference of squares is used as a formula in integral calculus. It can be derived in mathematically by the indefinite integration.
$\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx$
As per the difference of squares formula, the difference of squares of two terms can be factored mathematically.
$\implies$ $\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{(x+a)(x-a)}\,}dx$
The function in rational form represents a rational expression that consists of non-repeated linear factors in the denominator. It can be decomposed as the sum of two partial fractions.
$\implies$ $\displaystyle \int{\dfrac{1}{(x+a)(x-a)}\,}dx$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{A}{x+a}+\dfrac{B}{x-a}\Bigg)\,}dx$
Substitute $x = -a$, then $A$ $\,=\,$ $\dfrac{1}{-a-a}$ $\,=\,$ $\dfrac{1}{-2a}$ $\,=\,$ $-\dfrac{1}{2a}$
Substitute $x = a$, then $B$ $\,=\,$ $\dfrac{1}{a+a}$ $\,=\,$ $\dfrac{1}{2a}$
Now, replace the values of constants $A$ and $B$ in the expression, which consists of partial fractions.
$\implies$ $\displaystyle \int{\Bigg(\dfrac{A}{x+a}+\dfrac{B}{x-a}\Bigg)\,}dx$ $\,=\,$ $\displaystyle \int{\Bigg(\dfrac{\Big(-\dfrac{1}{2a}\Big)}{x+a}+\dfrac{\Big(\dfrac{1}{2a}\Big)}{x-a}\Bigg)\,}dx$
Now, simplify the algebraic expression in the rational form for evaluating the indefinite integration.
$=\,\,\,$ $\displaystyle \int{\Bigg(\dfrac{-1 \times \Big(\dfrac{1}{2a}\Big)}{x+a}+\dfrac{1 \times \Big(\dfrac{1}{2a}\Big)}{x-a}\Bigg)\,}dx$
In this mathematical expression,the factor $\dfrac{1}{2a}$ is common and it can be taken out from them.
$=\,\,\,$ $\displaystyle \int{\dfrac{1}{2a}\Bigg(\dfrac{-1}{x+a}+\dfrac{1}{x-a}\Bigg)\,}dx$
$=\,\,\,$ $\displaystyle \dfrac{1}{2a} \int{\Bigg(\dfrac{-1}{x+a}+\dfrac{1}{x-a}\Bigg)\,}dx$
$=\,\,\,$ $\displaystyle \dfrac{1}{2a} \int{\Bigg(\dfrac{1}{x-a}-\dfrac{1}{x+a}\Bigg)\,}dx$
$=\,\,\,$ $\displaystyle \dfrac{1}{2a}\Bigg(\int{\dfrac{1}{x-a}\,}dx-\int{\dfrac{1}{x+a}\,}dx\Bigg)$
Take $y = x-a$ and differentiate the equation with respect to $x$.
$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}{\, (x-a)}$
$\implies$ $\dfrac{dy}{dx} \,=\, \dfrac{d}{dx}{\, (x)}-\dfrac{d}{dx}{\, (a)}$
$\implies$ $\dfrac{dy}{dx} \,=\, 1-0$
$\implies$ $\dfrac{dy}{dx} \,=\, 1$
$\implies$ $dy \,=\, 1 \times dx$
$\implies$ $dy \,=\, dx$
$\,\,\, \therefore \,\,\,\,\,\,$ $dx \,=\, dy$
Take $z = x+a$ and differentiate the equation with respect to $x$.
$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}{\, (x+a)}$
$\implies$ $\dfrac{dz}{dx} \,=\, \dfrac{d}{dx}{\, (x)}+\dfrac{d}{dx}{\, (a)}$
$\implies$ $\dfrac{dz}{dx} \,=\, 1+0$
$\implies$ $\dfrac{dz}{dx} \,=\, 1$
$\implies$ $dz \,=\, 1 \times dx$
$\implies$ $dz \,=\, dx$
$\,\,\, \therefore \,\,\,\,\,\,$ $dx \,=\, dz$
Now, transform the terms in the expression in terms of $y$ and $z$.
$\implies$ $\displaystyle \dfrac{1}{2a}\Bigg(\int{\dfrac{1}{x-a}\,}dx-\int{\dfrac{1}{x+a}\,}dx\Bigg)$ $\,=\,$ $\displaystyle \dfrac{1}{2a}\Bigg(\int{\dfrac{1}{y}\,}dy-\int{\dfrac{1}{z}\,}dz\Bigg)$
Use the reciprocal rule of integration to evaluate integral of each multiplicative inverse of variable.
$= \,\,\,$ $\displaystyle \dfrac{1}{2a}\Bigg(\int{\dfrac{1}{y}\,}dy-\int{\dfrac{1}{z}\,}dz\Bigg)$
$= \,\,\,$ $\dfrac{1}{2a}\Big(\log_e{|y|}+c_1-\log_e{|z|}+c_2\Big)$
$= \,\,\,$ $\dfrac{1}{2a}\Big(\log_e{|y|}-\log_e{|z|}+(c_1+c_2)\Big)$
$= \,\,\,$ $\dfrac{1}{2a}\Big(\log_e{|y|}-\log_e{|z|}\Big)$ $+$ $\dfrac{1}{2a}(c_1+c_2)$
The difference of the logarithmic terms can be simplified by the quotient rule of logarithms.
$= \,\,\,$ $\dfrac{1}{2a}\Big(\log_e{\Bigg|\dfrac{y}{z}\Bigg|}\Big)$ $+$ $\dfrac{c_1+c_2}{2a}$
$= \,\,\,$ $\dfrac{1}{2a}\log_e{\Bigg|\dfrac{y}{z}\Bigg|}+c$
Actually, we have taken that $y = x-a$ and $z = x+a$ in the expression.
$= \,\,\,$ $\dfrac{1}{2a}\log_e{\Bigg|\dfrac{x-a}{x+a}\Bigg|}+c$
$\therefore \,\,\,$ $\displaystyle \int{\dfrac{1}{x^2-a^2}\,}dx \,=\, \dfrac{1}{2a}\ln{\Bigg|\dfrac{x-a}{x+a}\Bigg|}+c$
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