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

Lemma for Primitive of $\dfrac 1 {a^2 - x^2}$

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

Proof

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

\(\equiv\)

\(\ds \frac A {a - x} + \frac B {a + x}\)

Difference of Two Squares

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

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(\equiv\)

\(\ds A \paren {a + x} + B \paren {a - x}\)

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

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.

$\blacksquare$