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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.149$