Primitive of Reciprocal of x squared minus a squared/Logarithm Form 1

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Let $x \in \R$.

Then:

$\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}$


Proof

Case where $\size x < a$

Let $\size x < a$.

Then:

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

Let $\size x < a$.

Then:

\(\ds \int \frac {\d x} {x^2 - a^2}\) \(=\) \(\ds -\frac 1 a \artanh {\frac x a} + C\) Primitive of $\dfrac 1 {x^2 - a^2}$: $\artanh$ form
\(\ds \) \(=\) \(\ds -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} } } + C\) $\artanh \dfrac x a$ in Logarithm Form
\(\ds \) \(=\) \(\ds -\dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C\) Logarithm of Reciprocal


Case where $\size x > a$

Let $\size x > a$.

Then:

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

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

$\blacksquare$


Also see


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Primitive_of_Reciprocal_of_x_squared_minus_a_squared/Logarithm_Form_1&oldid=543687"