Primitive of Reciprocal of a squared minus x squared/Logarithm Form 2/Corollary

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Theorem

Let $a \in \R_{>0}$ be a strictly positive real constant.

Let $x \in \R$ such that $\size x \ne a$.


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


Proof

Let $z = b x$.

Then:

$\dfrac {\d x} {\d z} = \dfrac 1 b$

Hence:

\(\ds \int \frac {\d x} {a^2 - b^2 x^2}\) \(=\) \(\ds \int \dfrac 1 b \frac {\d z} {a^2 - z^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \dfrac 1 b \cdot \dfrac 1 {2 a} \ln \size {\dfrac {a + x} {a - x} } + C\) Primitive of $\dfrac 1 {a^2 - z^2}$: Logarithm Form $2$
\(\ds \) \(=\) \(\ds \dfrac 1 {2 a b} \ln \size {\dfrac {a + b x} {a - b x} } + C\) subtituting for $z$ and simplifying

$\blacksquare$


Sources

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