Primitive of Reciprocal of a squared minus x 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 $\size x \ne a$.


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


Corollary

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


Proof

From the $1$st logarithm form:

$\ds \int \frac {\d x} {a^2 - x^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C & : \size x < a\\ & \\ \dfrac 1 {2 a} \map \ln {\dfrac {x + a} {x - a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$


From Primitive of Reciprocal of a squared minus x squared: Logarithm Form: Lemma:

$\map \ln {\dfrac {a + x} {a - x} }$ is defined if and only if $\size x < a$
$\map \ln {\dfrac {x + a} {x - a} }$ is defined if and only if $\size x > a$


Let $\size x < a$.

Then $\map \ln {\dfrac {a + x} {a - x} }$ is defined.

We have that:

\(\ds \dfrac {a + x} {a - x}\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {a + x} {a - x}\) \(=\) \(\ds \size {\dfrac {a + x} {a - x} }\)

So the result holds for $\size x < a$.


Let $\size x > a$.

Then $\map \ln {\dfrac {x + a} {x - a} }$ is defined.

We have:

We have that:

\(\ds \dfrac {x + a} {x - a}\) \(=\) \(\ds -\dfrac {a + x} {a - x}\)
\(\ds \) \(<\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \dfrac {x + a} {x - a}\) \(=\) \(\ds \size {\dfrac {a + x} {a - x} }\)

The result follows.

$\blacksquare$


Also presented as

This result is also seen presented in the form:

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


Sources