Primitive of Reciprocal of a squared minus x 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:

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


Proof 1

Let $\size x > a$.

Then:

\(\ds \int \frac {\d x} {a^2 - x^2}\) \(=\) \(\ds \frac 1 a \arcoth {\frac x a} + C\) Primitive of $\dfrac 1 {a^2 - x^2}$: $\arcoth$ 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

$\blacksquare$


Proof 2

Let $\size x > a$.

Then:

\(\ds \int \frac {\d x} {a^2 - x^2}\) \(=\) \(\ds \int \frac {\d x} {\paren {a + x} \paren {a - x} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \int \frac {\d x} {2 a \paren {a + x} } + \int \frac {\d x} {2 a \paren {a - x} }\) Partial Fraction Expansion
\(\ds \) \(=\) \(\ds \int \frac {\d x} {2 a \paren {x + a} } - \int \frac {\d x} {2 a \paren {x - a} }\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a} \int \frac {\d x} {x + a} - \frac 1 {2 a} \int \frac {\d x} {x - a}\) 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


If $x > a$, then both $x + a > 0$ and $x - a > 0$.

So $\dfrac {x + a} {x - a} > 0$ and so:

$\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$


If $x < -a$, then both $x + a < 0$ and $x - a < 0$.

So again $\dfrac {x + a} {x - a} > 0$ and so:

$\ln \size {\dfrac {x + a} {x - a} } = \map \ln {\dfrac {x + a} {x - a} }$


Hence the result.

$\blacksquare$


Sources

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