Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2/Proof 2/Partial Fraction Expansion

Lemma for Primitive of Reciprocal of $\paren {x^2 - a^2}$

$\dfrac 1 {x^2 - a^2} \equiv \dfrac 1 {2 a \paren {x - a} } - \dfrac 1 {2 a \paren {x + a} }$

$\ds \frac 1 {\paren {x^2 - a^2} }$

$\equiv$

$\ds \frac A {x - a} + \frac B {x + a}$

$\text {(1)}: \quad$

$\ds \leadsto \ \ $

$\ds 1$

$\equiv$

$\ds A \paren {x + a} + B \paren {x - a}$

multiplying through by $\paren {x^2 - a^2}$

Setting $x = a$ in $(1)$:

$\ds A \cdot 2 a + B \cdot 0$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds A$

$=$

$\ds \frac 1 {2 a}$

Setting $x = -a$ in $(1)$:

$\ds A \cdot 0 + B \cdot \paren {-2 a}$

$=$

$\ds 1$

$\ds \leadsto \ \ $

$\ds B$

$=$

$\ds \frac {-1} {2 a}$

Summarising:

$\ds A$

$=$

$\ds \frac 1 {2 a}$

$\ds B$

$=$

$\ds \frac {-1} {2 a}$

Hence the result.

$\blacksquare$