Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1/size of x greater than a
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Theorem
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x > a$.
Then:
- $\ds \int \frac {\d x} {x^2 - a^2} = \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C$
Proof
Let $\size x > a$.
Then:
\(\ds \int \frac {\d x} {x^2 - a^2}\) | \(=\) | \(\ds -\frac 1 a \arcoth \frac x a + C\) | Primitive of $\dfrac 1 {x^2 - a^2}$: $\arcoth$ form | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C\) | $\arcoth \dfrac x a$ in Logarithm Form | |||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C\) | Reciprocal of Logarithm |
$\blacksquare$
Sources
- 1960: Margaret M. Gow: A Course in Pure Mathematics ... (previous) ... (next): Chapter $10$: Integration: $10.4$. Standard integrals: Standard Forms: $\text {(viii)}$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: General Rules of Integration: $14.40$
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.144$