Primitive of Reciprocal of x squared minus a squared/Logarithm Form 2
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Theorem
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
- $\ds \int \frac {\d x} {x^2 - a^2} = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C$
Proof 1
Let $x > a$.
Then:
\(\ds \int \frac {\d x} {x^2 - a^2}\) | \(=\) | \(\ds -\frac 1 a \arcoth {\frac x a} + C\) | Primitive of Reciprocal of $x^2 - a^2$ in $\arcoth$ 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C\) | Logarithm of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C\) | as $\dfrac {x - a} {x + a} > 0$ for $x > a$ |
Let $x < -a$.
Let $z = -x$.
Then:
- $\d x = -\d z$
and we then have:
\(\ds \int \frac {\d x} {x^2 - a^2}\) | \(=\) | \(\ds \int \frac {-\d z} {\paren {-z}^2 - a^2}\) | Integration by Substitution | |||||||||||
\(\ds \) | \(=\) | \(\ds -\int \frac {\d z} {z^2 - a^2}\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds -\frac 1 {2 a} \map \ln {\frac {z - a} {z + a} } + C\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \map \ln {\frac {z + a} {z - a} } + C\) | Logarithm of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \map \ln {\frac {-x + a} {-x - a} } + C\) | substituting $-x$ back for $z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \map \ln {\frac {\size {x - a} } {\size {x + a} } } + C\) | as $x < -a$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C\) |
The result follows.
$\blacksquare$
Proof 2
\(\ds \int \frac {\d x} {x^2 - a^2}\) | \(=\) | \(\ds \int \frac {\d x} {\paren {x - a} \paren {x + a} }\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d x} {2 a \paren {x - a} } - \int \frac {\d x} {2 a \paren {x + a} }\) | Partial Fraction Expansion | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \int \frac {\d x} {x - a} - \frac 1 {2 a} \int \frac {\d x} {x + a}\) | Primitive of Constant Multiple of Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {2 a} \ln \size {x - a} - \frac 1 {2 a} \ln \size {x + a} + C\) | Primitive of Reciprocal | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 {2 a} \ln \size {\dfrac {x - a} {x + a} } + C\) | Difference of Logarithms |
$\blacksquare$
Proof 3
From the $1$st logarithm form:
- $\ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\
& \\
\dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\
& \\ \text {undefined} & : \size x = a \end {cases}$
From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma:
- $\map \ln {\dfrac {x - a} {x + a} }$ is defined if and only if $\size x > a$
- $\map \ln {\dfrac {a - x} {a + x} }$ is defined if and only if $\size x < a$
Let $\size x > a$.
Then $\map \ln {\dfrac {x - a} {x + a} }$ is defined.
We have that:
\(\ds \dfrac {x - a} {x + a}\) | \(>\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {x - a} {x + a}\) | \(=\) | \(\ds \size {\dfrac {x - a} {x + a} }\) |
So the result holds for $\size x > a$.
Let $\size x < a$.
Then $\map \ln {\dfrac {a - x} {a + x} }$ is defined.
We have:
We have that:
\(\ds \dfrac {a - x} {a + x}\) | \(=\) | \(\ds -\dfrac {x - a} {x + a}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {a - x} {a + x}\) | \(=\) | \(\ds \size {\dfrac {x - a} {x + a} }\) |
The result follows.
$\blacksquare$
Sources
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- 1972: Frank Ayres, Jr. and J.C. Ault: Theory and Problems of Differential and Integral Calculus (SI ed.) ... (previous) ... (next): Chapter $25$: Fundamental Integration Formulas: $21$.
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- 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): Appendix $8$: Integrals