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

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

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 plus a squared

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

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 plus a squared

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

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}^{\left({n - 1}\right) + 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 \paren {x^2 + a^2}^{n - 1} } - \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}
      | 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 \paren {x^2 + a^2}^{n - 1} } - \frac 1 {a^2} \int \frac {\d x} {x^{m - 2} \paren {x^2 + a^2}^n}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over x squared minus a squared

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

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} }
      | 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 = {{Corollary|Primitive of Reciprocal}} as $z > 0$
}}
{{eqn | r = \frac 1 2 \map \ln {x^2 - a^2} + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over x squared minus a squared

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

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

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2} }
      | r = \int \paren {\frac x {a^2 \paren {x^2 - a^2} }  - \frac 1 {a^2 x} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \int \frac {\d x} x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \int \frac {x \rd x} {x^2 - a^2} - \frac 1 {a^2} \ln \size x + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^2} \paren {\frac 1 2 \map \ln {x^2 - a^2} } - \frac 1 {a^2} \ln \size x + C
      | c = Primitive of $\dfrac x {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \ln \size {x^2} + C
      | c = Logarithm of Power
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {x^2 - a^2} - \frac 1 {2 a^2} \map \ln {x^2} + C
      | c = as $x^2 > 0$
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2 - a^2} {x^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by x squared minus a squared

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

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

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 - a^2} }
      | r = \int \paren {\frac {-1} {a^2 x^3} - \frac 1 {a^4 x} + \frac x {a^4 \paren {x^2 - a^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 {x^2 -a^2} } + C
      | c = Primitive of $\dfrac x {x^2 - a^2}$
}}
{{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 {x^2 - a^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 x squared minus a squared squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {x^2 - a^2}^2}
      | r = \int \paren {\frac 1 {4 a^3 \paren {x + a} } - \frac 1 {4 a^3 \paren {x - a} } + \frac 1 {4 a^2 \paren {x + a}^2} + \frac 1 {4 a^2 \paren {x - a}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {4 a^3} \int \frac {\d x} {x + a} - \frac 1 {4 a^3} \int \frac {\d x} {x - a} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x - a}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {4 a^3} \ln \size {x + a} - \frac 1 {4 a^3} \ln \size {x - a} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x - a}^2} + C
      | c = Primitive of Function of $a x + b$ and Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {4 a^3} \ln \size {\frac {x + a} {x - a} } + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x - a}^2} + C
      | c = Difference of Logarithms
}}
{{eqn | r = \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^2} \int \frac {\d x} {\paren {x - a}^2} + C
      | c = Sign of Quotient of Factors of Difference of Squares
}}
{{eqn | r = \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } + \frac 1 {4 a^2} \frac {-1} {\paren {x + a} } + \frac 1 {4 a^2} \frac {-1} {\paren {x - a} } + C
      | c = Primitive of Function of $a x + b$ and Primitive of Power
}}
{{eqn | r = \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } - \paren {\frac 1 {4 a^2 \paren {x + a} } + \frac 1 {4 a^2 \paren {x - a} } } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } - \paren {\frac {\paren {x - a} + \paren {x + a} } {4 a^2 \paren {x + a} \paren {x - a} } } + C
      | c = simplifying
}}
{{eqn | r = \frac {-x} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {4 a^3} \map \ln {\frac {x + a} {x - a} } + C
      | c = Difference of Two Squares
}}
{{end-eqn}}
{{qed}}
Theorem: Chapman-Kolmogorov Equation

Let $X$ be a homogeneous Markov chain with $n$-step transition probability matrix:
:$\mathbf P^{\paren n} = \sqbrk { {p_{j k} }^{\paren n} }_{j, k \mathop \in S}$
where:
:${p_{j k} }^{\paren n} = \condprob {X_n = k} {X_0 = j} $ is the $n$-step transition probability.
Then:
:$\mathbf P^{\paren {n + m} } = \mathbf P^{\paren n} \mathbf P^{\paren m}$
or equivalently:
:$\ds {p_{i j} }^{\paren {n + m} } = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}$

Proof:
We consider the conditional probability on the {{LHS}}:
{{begin-eqn}}
{{eqn | l = \ds {p_{i j} }^{\paren {n + m} }
      | r = \map \Pr {X_{m + n} = j \mid X_0 = i}
}}
{{eqn | r = \condprob {\paren {\bigcup_{k \mathop \in S} \sqbrk {X_{n + m} = j, X_n = k} } } {X_0 = i}
}}
{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j, X_n = k} {X_0 = i}
      | c = {{Defof|Countable Additivity}}
}}
{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k, X_0 = i} \times \condprob {X_n = k} {X_0 = i}
      | c = Chain Rule for Probability
}}
{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{n + m} = j} {X_n = k} \times \condprob {X_n = k} {X_0 = i}
      | c = Markov Property
}}
{{eqn | r = \sum_{k \mathop \in S} \condprob {X_{0 + m} = j} {X_0 = k} \times \condprob {X_n = k} {X_0 = i}
      | c = Transition Probabilities of Homogeneous Markov Chain
}}
{{eqn | r = \sum_{k \mathop \in S} {p_{i k} }^{\paren n} {p_{k j} }^{\paren m}
}}
{{end-eqn}}
{{qed}}
{{Namedfor|Sydney Chapman|name2 = Andrey Nikolaevich Kolmogorov|cat = Chapman|cat2 = Kolmogorov}}
Category:Markov Chains
7ecrwp4h0pb6tytb1fq5gt38mg69e5i
Theorem: Primitive of x over x squared minus a squared squared

:$\ds \int \frac {x \rd x} {\paren {x^2 - a^2}^2} = \frac {-1} {2 \paren {x^2 - a^2} } + C$
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 = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = \int \frac {x \rd x} {\paren {x^2 - a^2}^2}
      | r = \int \frac {\d z} {2 z^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac {-1} {2 z} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 \paren {x^2 - a^2} } + C
      | c = substituting for $z$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over x squared minus a squared squared

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {\paren {x^2 - a^2}^2}
      | r = \int \frac {x^2 - a^2 + a^2} {\paren {x^2 - a^2}^2} \rd x
      | c = 
}}
{{eqn | r = \int \frac {x^2 - a^2} {\paren {x^2 - a^2}^2} \rd x + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int \frac {\d x} {x^2 - a^2} + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2}
      | c = simplification
}}
{{eqn | r = \frac 1 {2 a} \paren {\frac {x - a} {x + a} } + a^2 \int \frac {\d x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {2 a} \paren {\frac {x - a} {x + a} } + a^2 \paren {\frac {-x} {2 a^2 \paren {x^2 - a^2} } - \frac 1 {4 a^3} \paren {\frac {x - a} {x + a} } } + C
      | c = Primitive of $\dfrac 1 {\paren {x^2 - a^2}^2}$
}}
{{eqn | r = \frac {-x} {2 \paren {x^2 - a^2} } + \frac 1 {4 a} \map \ln {\frac {x - a} {x + a} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x cubed over x squared minus a squared squared

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

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

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 - a^2}^2}
      | r = \int \paren {\frac 1 {a^4 x} + \frac {-x} {a^4 \paren {x^2 - a^2} } + \frac x {a^2 \paren {x^2 - a^2}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {a^4} \ln \size x + \frac {-1} {a^4} \paren {\frac 1 2 \map \ln {x^2 - a^2} } + \frac 1 {a^2} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of $\dfrac x {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {a^4} \ln \size x  + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C
      | c = Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$
}}
{{eqn | r = \frac 1 {2 a^4} \map \ln {x^2} + \frac {-1} {2 a^4} \map \ln {x^2 - a^2} + \frac 1 {a^2} \paren {\frac {-1} {2 \paren {x^2 - a^2} } } + C
      | c = Logarithm of Power and $x^2 > 0$
}}
{{eqn | r = \frac {-1} {2 a^2 \paren {x^2 - a^2} } + \frac 1 {2 a^4} \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 x squared minus a squared squared

:$\ds \int \frac {\d x} {x^2 \paren {x^2 - a^2}^2} = \frac {-1} {a^4 x} - \frac x {2 a^4 \paren {x^2 - a^2} } + \frac 3 {4 a^5} \map \ln {\frac {x + a} {x - a} } + C$
for $x^2 > a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {x^2 - a^2}^2}
      | r = \int \paren {\frac 1 {a^4 x^2} + \frac 3 {4 a^5 \paren {x + a} } - \frac 3 {4 a^5 \paren {x - a} } + \frac 1 {4 a^4 \paren {x + a}^2} + \frac 1 {4 a^4 \paren {x - a}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^2} + \frac 3 {4 a^5} \int \frac {\d x} {x + a} - \frac 3 {4 a^5} \int \frac {\d x} {x - a} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x - a}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \frac {-1} x + \frac 3 {4 a^5} \int \frac {\d x} {x + a} - \frac 3 {4 a^5} \int \frac {\d x} {x - a} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x - a}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {x + a} - \frac 3 {4 a^5} \ln \size {x - a} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x + a}^2} + \frac 1 {4 a^4} \int \frac {\d x} {\paren {x - a}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {x + a} - \frac 3 {4 a^5} \ln \size {x - a} + \frac 1 {4 a^4} \frac {-1} {x + a} + \frac 1 {4 a^4} \frac {-1} {x - a} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \ln \size {\frac {x + a} {x - a} } + \frac {-1} {4 a^4 \paren { x + a} } + \frac {-1} {4 a^4 \paren {x - a} } + C
      | c = Difference of Logarithms
}}
{{eqn | r = \frac {-1} {a^4 x} + \frac 3 {4 a^5} \map \ln {\frac {x + a} {x - a} } + \frac {-1} {4 a^4 \paren { x + a} } + \frac {-1} {4 a^4 \paren {x - a} } + C
      | c = Sign of Quotient of Factors of Difference of Squares
}}
{{eqn | r = \frac {-1} {a^4 x} - \frac x {2 a^4 \paren {x^2 - a^2} } + \frac 3 {4 a^5} \map \ln {\frac {x + a} {x - a} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x cubed by x squared minus a squared squared

:$\ds \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2} = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C$
for $x^2 > a^2$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {x^2 - a^2}^2}
      | r = \int \paren {\frac 1 {a^4 x^3} + \frac 2 {a^6 x} - \frac {2 x} {a^6 \paren {x^2 - a^2} } + \frac x {a^4 \paren {x^2 - a^2}^2} } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {a^4} \int \frac {\d x} {x^3} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^4} \frac {-1} {2 x^2} + \frac 2 {a^6} \int \frac {\d x} x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \int \frac {x \rd x} {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 2 {a^6} \frac 1 2 \map \ln {x^2 - a^2} + \frac 1 {a^4} \int \frac {x \rd x} {\paren {x^2 - a^2}^2} + C
      | c = Primitive of $\dfrac x {x^2 - a^2}$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 2 {a^6} \ln \size x - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C
      | c = Primitive of $\dfrac x {\paren {x^2 - a^2}^2}$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} + \frac 1 {a^6} \map \ln {x^2} - \frac 1 {a^6} \map \ln {x^2 - a^2} + \frac 1 {a^4} \frac {-1} {2 \paren {x^2 - a^2} }  + C
      | c = Logarithm of Power and $x^2 > 0$
}}
{{eqn | r = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^4 \paren {x^2 - a^2} } + \frac 1 {a^6} \map \ln {\frac {x^2} {x^2 - a^2} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}