Primitive of Reciprocal of x by x squared minus a squared

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Theorem

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

for $x^2 > a^2$.


Proof

\(\displaystyle \int \frac {\d x} {x \paren {x^2 - a^2} }\) \(=\) \(\displaystyle \int \paren {\frac x {a^2 \paren {x^2 - a^2} } - \frac 1 {a^2 x} } \rd x\) Partial Fraction Expansion
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} x\) Linear Combination of Integrals
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \ln \size x + C\) Primitive of Reciprocal
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 - a^2} } - \frac 1 {a^2} \ln \size x + C\) Primitive of $\dfrac x {x^2 - a^2}$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \ln \size {x^2} + C\) Logarithm of Power
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \map \ln {x^2} + C\) as $x^2 > 0$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C\) Difference of Logarithms

$\blacksquare$


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Sources