avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Reciprocal of Power of x squared minus a squared

:$\ds \int \frac {\d x} {\paren {x^2 - a^2}^n} = \frac {-x} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.

Proof:
Aiming for an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \ \frac {\d u} {\d x} \rd x$
in order to use the technique of Integration by Parts, let:
{{begin-eqn}}
{{eqn | l = v
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d v} {\d x}
      | r = 1
      | c = Power Rule for Derivatives
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac 1 {\paren {x^2 - a^2}^{n - 1} }
      | c = 
}}
{{eqn | r = \paren {x^2 - a^2}^{-\paren {n - 1} }
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = -\paren {n - 1} \paren {2 x} \paren {x^2 - a^2}^{-\paren {n - 1} - 1}
      | c = Power Rule for Derivatives and Chain Rule for Derivatives
}}
{{eqn | r = -2 \paren {n - 1} x \paren {x^2 - a^2}^{-n}
      | c = 
}}
{{eqn | r = \frac {-\paren {2 n - 2} x} {\paren {x^2 - a^2}^n}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} }
      | r = \frac 1 {\paren {x^2 - a^2}^{n - 1} } x - \int x \frac {-\paren {2 n - 2} x} {\paren {x^2 - a^2}^n} \rd x
      | c = Integration by Parts
}}
{{eqn | r = \frac x {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2} {\paren {x^2 - a^2}^n} \rd x
      | c = simplifying
}}
{{eqn | r = \frac x {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2 - a^2 + a^2} {\paren {x^2 - a^2}^n} \rd x
      | c = 
}}
{{eqn | r = \frac x {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {x^2 - a^2} {\paren {x^2 - a^2}^n} \rd x + \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 - a^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac x {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 - a^2}^n} 
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {1 - \paren {2 n - 2} } \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} }
      | r = \frac x {\paren {x^2 - a^2}^{n - 1} } + \paren {2 n - 2} a^2 \int \frac {\d x} {\paren {x^2 - a^2}^n}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {x^2 - a^2}^n}
      | r = \frac {-x} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac {2 n - 3} {\paren {2 n - 2} a^2} \int \frac {\d x} {\paren {x^2 - a^2}^{n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over Power of x squared minus a squared

:$\ds \int \frac {x \rd x} {\paren {x^2 - a^2}^n} = \frac {-1} {2 \paren {n - 1} \paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2 - a^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {\paren {x^2 - a^2}^n}
      | r = \int \frac {\d z} {2 z^n}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int z^{-n} \rd z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 2 \frac {z^{-n + 1} } {-n + 1} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 \paren {n - 1} z^{n - 1} } + C
      | c = simplifying
}}
{{eqn | r = \frac {-1} {2 \paren {n - 1} \paren {x^2 - a^2}^{n - 1} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by Power of x squared minus a squared

:$\ds \int \frac {\d x} {x \paren {x^2 - a^2}^n} = \frac {-1} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x \paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = x^2 - a^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \paren {x^2 - a^2}^n}
      | r = \int \frac {\d z} {2 x^2 z^n}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 2 \int \frac {\d z} {\paren {z + a^2} z^n}
      | c = Primitive of Constant Multiple of Function and substituting for $x^2$
}}
{{end-eqn}}
From Primitive of $x^m \paren {a x + b}^n$: Increment of Power of $x$:
:$\ds \int x^m \paren {a x + b}^n \rd x = \frac {x^{m + 1} \paren {a x + b}^{n + 1} } {\paren {m + 1} b} - \frac {\paren {m + n + 2} a} {\paren {m + 1} b} \int x^{m + 1} \paren {a x + b}^n \rd x$
Let:
{{begin-eqn}}
{{eqn | l = a
      | o = :=
      | r = 1
      | c = 
}}
{{eqn | l = b
      | o = :=
      | r = a^2
      | c = 
}}
{{eqn | l = x
      | o = :=
      | r = z
      | c = 
}}
{{eqn | l = m
      | o = :=
      | r = -n
      | c = 
}}
{{eqn | l = n
      | o = :=
      | r = -1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2}^n}
      | r = \frac 1 2 \int \frac {\d z} {\paren {z + a^2} z^n}
      | c = 
}}
{{eqn | r = \frac 1 2 \paren {\frac {z^{-n + 1} \paren {z + a^2}^{-1 + 1} } {\paren {-n + 1} a^2} - \frac {\paren {-n + \paren {-1} + 2} } {\paren {-n + 1} a^2} \int z^{-n + 1} \paren {z + a^2}^{-1} \rd z}
      | c = 
}}
{{eqn | r = \frac {-1} {2 \paren {n - 1} a^2 z^{n-1} } - \frac 1 {2 a^2} \int \frac {\d z} {z^{n - 1} \paren {z + a^2} }
      | c = simplification
}}
{{eqn | r = \frac {-1} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac 1 {2 a^2} \int \frac {2 x \rd x} {x^2 \paren {x^2 - a^2}^{n - 1} }
      | c = substituting for $z$ and $\d z$
}}
{{eqn | r = \frac {-1} {2 \paren {n - 1} a^2 \paren {x^2 - a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x \paren {x^2 - a^2}^{n - 1} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x over Power of x squared minus a squared

:$\ds \int \frac {x^m \rd x} {\paren {x^2 - a^2}^n} = \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^{n - 1} } + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}$
for $x^2 > a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^m \rd x} {\paren {x^2 - a^2}^n}
      | r = \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {x^2 - a^2}^n}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2 + a^2} \rd x} {\paren {x^2 - a^2}^n}
      | c = 
}}
{{eqn | r = \int \frac {x^{m - 2} \paren {x^2 - a^2} \rd x} {\paren {x^2 - a^2}^n} + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^{n - 1} } + a^2 \int \frac {x^{m - 2} \rd x} {\paren {x^2 - a^2}^n}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of x by Power of x squared minus a squared

:$\ds \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }$
for $x^2 > a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
      | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{n - 1} \paren {x^2 - a^2} }
      | c = 
}}
{{eqn | r = \int \frac {\paren {x^2 - a^2} \rd x} {x^m \paren {x^2 - a^2}^{\paren {n - 1} + 1} }
      | c = 
}}
{{eqn | r = \int \frac {x^2 \rd x} {x^m \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n}
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = a^2 \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} 
      | r = \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
      | c = changing sides
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^m \paren {x^2 - a^2}^n} 
      | r = \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 - a^2}^n} - \frac 1 {a^2} \int \frac {\d x} {x^m \paren {x^2 - a^2}^{n - 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Sign of Quotient of Factors of Difference of Squares

Let $a, b \in \R$ such that $a \ne b$.
Then
:$\map \sgn {a^2 - b^2} = \map \sgn {\dfrac {a + b} {a - b} } = \map \sgn {\dfrac {a - b} {a + b} }$
where $\sgn$ denotes the signum of a real number.

Proof:
{{begin-eqn}}
{{eqn | l = \map \sgn {\frac {a - b} {a + b} }
      | r = \map \sgn {a - b} \frac 1 {\map \sgn {a + b} }
      | c = Signum Function is Completely Multiplicative
}}
{{eqn | r = \map \sgn {a - b} \map \sgn {a + b}
      | c = Signum Function of Reciprocal
}}
{{eqn | r = \map \sgn {\paren {a - b} \paren {a + b} }
      | c = Signum Function is Completely Multiplicative
}}
{{eqn | r = \map \sgn {a^2 - b^2}
      | c = Difference of Two Squares
}}
{{eqn | r = \map \sgn {\paren {a + b} \paren {a - b} }
      | c = Difference of Two Squares
}}
{{eqn | r = \map \sgn {a + b} \map \sgn {a - b}
      | c = Signum Function is Completely Multiplicative
}}
{{eqn | r = \map \sgn {a + b} \frac 1 {\map \sgn {a - b} }
      | c = Signum Function of Reciprocal
}}
{{eqn | r = \map \sgn {\frac {a + b} {a - b} }
      | c = Signum Function is Completely Multiplicative
}}
{{end-eqn}}
{{qed}}
Category:Real Analysis
Category:Signum Function
ss7k87h3t6ty6ypqnd7cqi9f1xktmko
Theorem: Signum Function is Completely Multiplicative

The signum function on the set of real numbers is a completely multiplicative function:
:$\forall x, y \in \R: \map \sgn {x y} = \map \sgn x \map \sgn y$

Proof:
Let $x = 0$ or $y = 0$.
Then:
{{begin-eqn}}
{{eqn | l = x y
      | r = 0
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \map \sgn {x y}
      | r = 0
      | c = 
}}
{{end-eqn}}
and either $\map \sgn x = 0$ or $\map \sgn y = 0$ and so:
{{begin-eqn}}
{{eqn | l = \map \sgn x \map \sgn y
      | r = 0
      | c = 
}}
{{eqn | r = \map \sgn {x y}
      | c = from $(1)$ above
}}
{{end-eqn}}
{{qed|lemma}}
Let $x > 0$ and $y > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \sgn x
      | r = 1
      | c = 
}}
{{eqn | lo= \land
      | l = \map \sgn y
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \sgn x \map \sgn y
      | r = 1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x y
      | o = >
      | r = 0
      | c = 
}}
{{eqn | l = \map \sgn {x y}
      | r = 1
      | c ={{Defof|Signum Function}}
}}
{{eqn | r = \map \sgn x \map \sgn y
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Let $x < 0$ and $y < 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \sgn x
      | r = -1
      | c = 
}}
{{eqn | lo= \land
      | l = \map \sgn y
      | r = -1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \sgn x \map \sgn y
      | r = 1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x y
      | o = >
      | r = 0
      | c = 
}}
{{eqn | l = \map \sgn {x y}
      | r = 1
      | c = {{Defof|Signum Function}}
}}
{{eqn | r = \map \sgn x \map \sgn y
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Let $x < 0$ and $y > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \map \sgn x
      | r = -1
      | c = 
}}
{{eqn | lo= \land
      | l = \map \sgn y
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \sgn x \map \sgn y
      | r = -1
      | c = 
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | l = x y
      | o = <
      | r = 0
      | c = 
}}
{{eqn | l = \map \sgn {x y}
      | r = -1
      | c = {{Defof|Signum Function}}
}}
{{eqn | r = \map \sgn x \map \sgn y
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
The same argument, {{mutatis}}, covers the case where $x > 0$ and $y < 0$.
{{qed}}
Category:Signum Function
Category:Completely Multiplicative Functions
alwj2om9na0p1552b7an6me5oidkol1
Theorem: Signum Function is Quotient of Number with Absolute Value

Let $x \in \R_{\ne 0}$ be a non-zero real number.
Then:
:$\map \sgn x = \dfrac x {\size x} = \dfrac {\size x} x$
where:
:$\map \sgn x$ denotes the signum function of $x$
:$\size x$ denotes the absolute value of $x$.

Proof:
Let $x \in \R_{\ne 0}$.
Then either $x > 0$ or $x < 0$.
Let $x > 0$.
Then:
{{begin-eqn}}
{{eqn | l = \frac x {\size x}
      | r = \frac x x
      | c = {{Defof|Absolute Value}}, as $x > 0$
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = \map \sgn x
      | c = {{Defof|Signum Function}}, as $x > 0$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \frac {\size x} x
      | r = \frac x x
      | c = {{Defof|Absolute Value}}, as $x > 0$
}}
{{eqn | r = 1
      | c = 
}}
{{eqn | r = \map \sgn x
      | c = {{Defof|Signum Function}}, as $x > 0$
}}
{{end-eqn}}
{{qed|lemma}}
Let $x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = \frac x {\size x}
      | r = \frac x {-x}
      | c = {{Defof|Absolute Value}}, as $x < 0$
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | r = \map \sgn x
      | c = {{Defof|Signum Function}}, as $x < 0$
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \frac {\size x} x
      | r = \frac {-x} x
      | c = {{Defof|Absolute Value}}, as $x < 0$
}}
{{eqn | r = -1
      | c = 
}}
{{eqn | r = \map \sgn x
      | c = {{Defof|Signum Function}}, as $x < 0$
}}
{{end-eqn}}
{{qed}}
Category:Signum Function
Category:Absolute Value Function
edyk3jk00xbgeqk3e2vrhz1wzgu6dju
Theorem: Completely Multiplicative Function of Quotient

Let $f: \R \to \R$ be a completely multiplicative function.
Then:
:$\forall x, y \in \R, y \ne 0: \map f {\dfrac x y} = \dfrac {\map f x} {\map f y}$
whenever $\map f y \ne 0$.

Proof:
Let $z = \dfrac x y$.
Then:
{{begin-eqn}}
{{eqn | l = z
      | r = \dfrac x y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y z
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map f x
      | r = \map f y \map f z
      | c = {{Defof|Completely Multiplicative Function}}
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f x} {\map f y}
      | r = \map f z
      | c = dividing by $\map f y$
}}
{{eqn | ll= \leadsto
      | l = \frac {\map f x} {\map f y}
      | r = \map f {\frac x y}
      | c = Definition of $z$
}}
{{end-eqn}}
{{qed}}
Category:Multiplicative Functions
c9t6rb0jneowm1mog3zviifwzqizhpe
Theorem: Signum Function of Reciprocal

Let $x \in \R$ such that $x \ne 0$.
Then:
:$\map \sgn x = \map \sgn {\dfrac 1 x}$
where $\map \sgn x$ denotes the signum of $x$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \sgn x
      | r = 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = >
      | r = 0
      | c = {{Defof|Signum Function}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \frac 1 x
      | o = >
      | r = 0
      | c = Reciprocal of Strictly Positive Real Number is Strictly Positive
}}
{{eqn | ll= \leadstoandfrom
      | l = \map \sgn {\dfrac 1 x}
      | r = 1
      | c = {{Defof|Signum Function}}
}}
{{end-eqn}}
{{qed}}
Category:Signum Function
3fktxynn8ed1viz3c18ts1pma7lwj8f
Theorem: Primitive of x over a squared minus x squared

:$\ds \int \frac {x \rd x} {a^2 - x^2} = -\frac 1 2 \map \ln {a^2 - x^2} + C$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a^2 - x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = -2 x
      | c = Power Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x \paren {a^2 - x^2} }
      | r = \int \frac {\d z} {-2 z}
      | c = Integration by Substitution
}}
{{eqn | r = -\frac 1 2 \int \frac {\d z} z
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 2 \ln z + C
      | c = Primitive of Reciprocal: Corollary as $z > 0$
}}
{{eqn | r = -\frac 1 2 \, \map \ln {a^2 - x^2} + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Linear Combination of Laplace Transforms

Then:
:$\laptrans {\lambda \, \map f t + \mu \, \map g t} = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}$
everywhere all the above expressions are defined.

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {\lambda \, \map f t + \mu \, \map g t}
      | r = \int_0^{\to +\infty} e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t
      | c = {{Defof|Laplace Transform}}
}}
{{eqn | r = \lim_{A \mathop \to +\infty} \paren {\int_0^A e^{-s t} \paren {\lambda \, \map f t + \mu \, \map g t} \rd t}
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lim_{A \mathop \to +\infty} \paren {\lambda \int_0^A e^{-s t} \map f t \rd t + \mu \int_0^A e^{-s t} \map g t \rd t}
      | c = distributing $e^{-s t}$, Linear Combination of Complex Integrals
}}
{{eqn | r = \lambda \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map f t \rd t + \mu \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} \map g t \rd t
      | c = Combination Theorem for Limits at Infinity
}}
{{eqn | r = \lambda \int_0^{\to +\infty} e^{-s t} \map f t \rd t + \mu \int_0^{\to +\infty} e^{-st} \map g t \rd t
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}
      | c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over a squared minus x squared

:$\ds \int \frac {x^3 \rd x} {a^2 - x^2} = -\frac {x^2} 2 - \frac {a^2} 2 \map \ln {a^2 - x^2} + C$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = \int \frac {x^3 \rd x} {a^2 - x^2}
      | r = \int \frac {x \paren {x^2 - a^2 + a^2} } {a^2 - x^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {-x \paren {a^2 - x^2} } {a^2 - x^2} \rd x + \int \frac {a^2 x} {a^2 - x^2} \rd x
      | c = 
}}
{{eqn | r = -\int x \rd x + a^2 \int \frac {x \rd x} {a^2 - x^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac {x^2} 2 + a^2 \int \frac {x \rd x} {a^2 - x^2} + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac {x^2} 2 + a^2 \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + C
      | c = Primitive of Reciprocal of $\dfrac x {a^2 - x^2}$
}}
{{eqn | r = -\frac {x^2} 2 - \frac {a^2} 2 \map \ln {a^2 - x^2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by a squared minus x squared

:$\ds \int \frac {\d x} {x \paren {a^2 - x^2} } = \frac 1 {2 a^2} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {a^2 - x^2} }
      | r = \int \paren {\frac 1 {a^2 x} + \frac x {a^2 \paren {a^2 - x^2} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} x + \frac 1 {a^2} \int \frac {x \rd x} {a^2 - x^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \ln \size x + \frac 1 {a^2} \int \frac {x \rd x} {a^2 - x^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^2} \ln \size x + \frac 1 {a^2} \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + C
      | c = Primitive of $\dfrac x {a^2 - x^2}$
}}
{{eqn | r = \frac 1 {2 a^2} \ln \size {x^2} - \frac 1 {2 a^2} \map \ln {a^2 - x^2} + C
      | c = Logarithm of Power
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {x^2} - \frac 1 {2 a^2} \map \ln {a^2 - x^2} + C
      | c = Square of Real Number is Non-Negative
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 - a^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by a squared minus x squared

:$\ds \int \frac {\d x} {x^2 \paren {a^2 - x^2} } = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a^2 - x^2} }
      | r = \int \paren {\frac 1 {a^2 \paren {a^2 - x^2} } + \frac 1 {a^2 x^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} + \frac 1 {a^2} \int \frac {\d x} {x^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {a^2 - x^2} - \frac 1 {a^2 x} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {\frac {a + x} {a - x} } } - \frac 1 {a^2 x} + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$
}}
{{eqn | r = \frac {-1} {a^2 x} + \frac 1 {2 a^3} \map \ln {\frac {a + x} {a - x} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by a squared minus x squared

:$\ds \int \frac {\d x} {x^3 \paren {a^2 - x^2} } = \frac {-1} {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a^2 - x^2} }
      | r = \int \paren {\frac 1 {a^2 x^3} + \frac 1 {a^4 x} + \frac x {a^4 \paren {a^2 - x^2} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} {x^3} + \frac 1 {a^4} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {x \rd x} {x^2 - a^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-1} {2 a^2 x^2} - \frac 1 {a^4} \int \frac {\d x} x + \frac 1 {a^4} \int \frac {x \rd x} {x^2 - a^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a^2 x^2} + \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \int \frac {x \rd x} {x^2 - a^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-1} {2 a^2 x^2} + \frac 1 {a^4} \ln \size x + \frac 1 {a^4} \paren {-\frac 1 2 \map \ln {a^2 - x^2} } + C
      | c = Primitive of $\dfrac x {a^2 - x^2}$
}}
{{eqn | r = \frac {-1} {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {x^2} - \frac 1 {2 a^4} \map \ln {a^2 - x^2} + C
      | c = Logarithm of Power
}}
{{eqn | r = \frac {-1} {2 a^2 x^2} + \frac 1 {2 a^4} \map \ln {\frac {x^2} {a^2 - x^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a squared minus x squared squared

:$\ds \int \frac {\d x} {\paren {a^2 - x^2}^2} = \frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {4 a^3} \map \ln {\frac {a + x} {a - x} } + C$
for $x^2 < a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a^2 - x^2}^2}
      | r = \int \paren {\frac 1 {4 a^2} \paren {\dfrac 1 {\paren {a - x}^2} + \frac 1 {\paren {a + x}^2} + \frac 1 {a \paren {a + x} } + \frac 1 {a \paren {a - x} } } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \int \paren {\frac 1 {4 a^2} \paren {\frac 1 {\paren {a - x}^2} + \frac 1 {\paren {a + x}^2} + \frac 2 {\paren {a^2 - x^2} } } } \rd x
      | c = Reciprocal of Difference of Squares as Sum of Reciprocals: $\dfrac 2 {a^2 - x^2} = \dfrac 1 {a \paren {a + x} } + \dfrac 1 {a \paren {a - x} }$
}}
{{eqn | r = \frac 1 {4 a^2} \paren {\int \frac {\d x} {\paren {a - x}^2} + \int \frac {\d x} {\paren {a + x}^2} } + \frac 1 {2 a^2} \int \frac {\d x} {a^2 - x^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {4 a^2} \paren {\frac 1 {a - x} - \frac 1 {a + x} } + \frac 1 {2 a^2} \int \frac {\d x} {a^2 - x^2}
      | c = Primitive of Function of $a x + b$ and Primitive of Power
}}
{{eqn | r = \frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^2} \int \frac {\d x} {a^2 - x^2}
      | c = Reciprocal of Difference of Squares as Difference of Reciprocals: $\dfrac x {a^2 - x^2} = \dfrac 1 {2 \paren {a + x} } - \dfrac 1 {2 \paren {a - x} }$
}}
{{eqn | r = \frac x {2 a^2 \paren {a^2 - x^2} } + \frac 1 {2 a^2} \paren {\dfrac 1 {2 a} \map \ln {\frac {a + x} {a - x} } } + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Primitive of x over a squared minus x squared squared

:$\ds \int \frac {x \rd x} {\paren {a^2 - x^2}^2} = \frac 1 {2 \paren {a^2 - x^2} } + C$
for $x^2 < a^2$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a^2 - x^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = -2 x
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {\paren {a^2 - x^2}^2}
      | r = \int \frac {\d z} {-2 z^2}
      | c = Integration by Substitution
}}
{{eqn | r = -\frac 1 2 \int \frac {\d z} {z^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 2 \paren {\frac {-1} z} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac 1 {2 \paren {a^2 - x^2} } + C
      | c = substituting for $z$ and simplifying
}}
{{end-eqn}}
{{qed}}