Primitive of x over x squared minus a squared

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Theorem

$\displaystyle \int \frac {x \rd x} {x^2 - a^2} = \frac 1 2 \map \ln {x^2 - a^2} + C$

for $x^2 > a^2$.


Proof

Let:

\(\displaystyle z\) \(=\) \(\displaystyle x^2 - a^2\)
\(\displaystyle \implies \ \ \) \(\displaystyle \frac {\d z} {\d x}\) \(=\) \(\displaystyle 2 x\) Power Rule for Derivatives
\(\displaystyle \implies \ \ \) \(\displaystyle \int \frac {\d x} {x \paren {x^2 - a^2} }\) \(=\) \(\displaystyle \int \frac {\d z} {2 z}\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \int \frac {\d z} z\) Primitive of Constant Multiple of Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \ln z + C\) Primitive of Reciprocal: Corollary as $z > 0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 2 \map \ln {x^2 - a^2} + C\) substituting for $z$

$\blacksquare$


Also see


Sources