Primitives involving x squared minus a squared
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Theorem
This page gathers together the primitives of some expressions involving $x^2 - a^2$.
Primitive of Reciprocal of $x^2 - a^2$: $\coth^{-1}$ form
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $\size x > a$.
Then:
- $\ds \int \frac {\d x} {x^2 - a^2} = -\frac 1 a \coth^{-1} {\frac x a} + C$
Primitive of Reciprocal of $x^2 - a^2$: Logarithm Form
Let $a \in \R_{>0}$ be a strictly positive real constant.
Let $x \in \R$ such that $\size x \ne a$.
$1$st Logarithm Form
- $\ds \int \frac {\d x} {x^2 - a^2} = \begin {cases} \dfrac 1 {2 a} \map \ln {\dfrac {a - x} {a + x} } + C & : \size x < a\\
& \\ \dfrac 1 {2 a} \map \ln {\dfrac {x - a} {x + a} } + C & : \size x > a \\ & \\ \text {undefined} & : \size x = a \end {cases}$
$2$nd Logarithm Form
- $\ds \int \frac {\d x} {x^2 - a^2} = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C$
Primitive $x$ over $x^2 - a^2$
- $\ds \int \frac {x \rd x} {x^2 - a^2} = \frac 1 2 \map \ln {x^2 - a^2} + C$
for $x^2 > a^2$.
Primitive $x^2$ over $x^2 - a^2$
Theorem
Logarithm Form
- $\ds \int \frac {x^2 \rd x} {x^2 - a^2} = x + \frac a 2 \map \ln {\frac {x - a} {x + a} } + C$
for $x^2 > a^2$.
Inverse Hyperbolic Cotangent Form
- $\ds \int \frac {x^2 \rd x} {x^2 - a^2} = x - a \coth^{-1} \frac x a + C$
for $x^2 > a^2$.
Also see
Primitive $x^3$ over $x^2 - a^2$
- $\ds \int \frac {x^3 \rd x} {x^2 - a^2} = \frac {x^2} 2 + \frac {a^2} 2 \map \ln {x^2 - a^2} + C$
for $x^2 > a^2$.
Primitive of Reciprocal of $x \paren {x^2 - a^2}$
- $\ds \int \frac {\d x} {x \paren {x^2 - a^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C$
for $x^2 > a^2$.
Primitive of Reciprocal of $x^2 \paren {x^2 - a^2}$
- $\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2} } = \frac 1 {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {x - a} {x + a} } + C$
for $x^2 > a^2$.
Primitive of Reciprocal of $x^3 \paren {x^2 - a^2}$
- $\ds \int \frac {\d x} {x^3 \paren {x^2 - a^2} } = \frac 1 {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {\frac {x^2 - a^2} {x^2} } + C$
for $x^2 > a^2$.