Primitive of Reciprocal of x squared by x squared minus a squared

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Theorem

$\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2} } = \frac 1 {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {x - a} {x + a} } + C$

for $x^2 > a^2$.


Proof

\(\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2} }\) \(=\) \(\ds \int \paren {\frac 1 {a^2 \paren {x^2 - a^2} } - \frac 1 {a^2 x^2} } \rd x\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} {x^2}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \int \frac {\d x} {x^2 - a^2} + \frac 1 {a^2 x} + C\) Primitive of Power
\(\ds \) \(=\) \(\ds \frac 1 {a^2} \paren {\frac 1 2 \map \ln {\frac {x - a} {x + a} } } + \frac 1 {a^2 x} + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {x - a} {x + a} } + C\) simplifying

$\blacksquare$


Also see


Sources

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