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

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c} } = \frac 1 {2 c} \ln \size {\frac {x^2} {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {a x^2 + b x + c}$

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

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } = \frac b {2 c^2} \ln \size {\frac {a x^2 + b x + c} {x^2} } - \frac 1 {c x}  + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} }
}}
{{eqn | r = \int \paren {\frac {-b} {c^2 x} + \frac 1 {c x^2} + \frac {a b x + b^2 - a c} {c^2 \paren {a x^2 + b x + c} } } \rd x
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac {-b} {c^2} \int \frac {\d x} x + \frac 1 c \int \frac {\d x} {x^2} + \frac {a b} {c^2} \int \frac {x \rd x} {a x^2 + b x + c} + \frac {b^2 - a c} {c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac {-b} {c^2} \ln \size x + \frac 1 c \int \frac {\d x} {x^2} + \frac {a b} {c^2} \int \frac {x \rd x} {a x^2 + b x + c} + \frac {b^2 - a c} {c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac {-b} {c^2} \ln \size x - \frac 1 {c x} + \frac {a b} {c^2} \int \frac {x \rd x} {a x^2 + b x + c} + \frac {b^2 - a c} {c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Primitive of Power
}}
{{eqn | r = \frac {-b} {c^2} \ln \size x - \frac 1 {c x} + \frac {a b} {c^2} \paren {\frac 1 {2 a} \ln \size {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c} } + \frac {b^2 - a c} {c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Primitive of $\dfrac x {a x^2 + b x + c}$
}}
{{eqn | r = \frac {-b} {c^2} \ln \size x - \frac 1 {c x} + \frac b {2 c^2} \ln \size {a x^2 + b x + c} - \frac {b^2} {2 c^2} \int \frac {\d x} {a x^2 + b x + c} + \frac {2 b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = multiplying out
}}
{{eqn | r = \frac {-b} {c^2} \ln \size x - \frac 1 {c x} + \frac b {2 c^2} \ln \size {a x^2 + b x + c} + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = simplifying
}}
{{eqn | r = \frac {-b} {2 c^2} \ln \size {x^2} - \frac 1 {c x} + \frac b {2 c^2} \ln \size {a x^2 + b x + c} + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Logarithm of Power
}}
{{eqn | r = \frac b {2 c^2} \ln \size {\frac {a x^2 + b x + c} {x^2} } - \frac 1 {c x}  + \frac {b^2 - 2 a c} {2 c^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of Power of x by a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^n \paren {a x^2 + b x + c} } = \frac {-1} {\paren {n - 1} c x^{n - 1} } - \frac b c \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }
      | r = \int \frac {x^{-n + 2} \rd x} {a x^2 + b x + c}
      | c = 
}}
{{eqn | r = \frac {x^{-n + 1} } {\paren {-n + 1} a} - \frac b a \int \frac {x^{-n + 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{-n} \rd x} {a x^2 + b x + c}
      | c = Primitive of $\dfrac {x^m} {a x^2 + b x + c}$
}}
{{eqn | ll= \leadsto
      | l = \frac c a \int \frac {\d x} {x^n \paren {a x^2 + b x + c} }
      | r = \frac {-1} {\paren {n - 1} a x^{n - 1} } - \frac b a \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }
      | c = rearranging
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {x^n \paren {a x^2 + b x + c} }
      | r = \frac {-1} {\paren {n - 1} c x^{n - 1} } - \frac b c \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }
      | c = multiplying by $\dfrac a c$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of square of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {\paren {a x^2 + b x + c}^2} = \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = 2 a x + b
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a
      | c = Derivative of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | r = \int \paren {\frac {4 a} {\paren {2 a x + b}^2 + 4 a c - b^2} }^2 \rd x
      | c = Completing the Square
}}
{{eqn | r = \int \frac {\paren {4 a}^2} {\paren {z^2 + 4 a c - b^2}^2} \frac {\d z} {2 a}
      | c = Integration by Substitution
}}
{{eqn | r = 8 a \int \frac {\d z} {\paren {z^2 + 4 a c - b^2}^2}
      | c = simplifying
}}
{{end-eqn}}
Let $u = z^2$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = z^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d z}
      | r = 2 z
      | c = Derivative of Power
}}
{{eqn | ll= \leadsto
      | l = 8 a \int \frac {\d z} {\paren {z^2 + 4 a c - b^2}^2}
      | r = 8 a \int \frac {\d u} {2 \sqrt u \paren {u + 4 a c - b^2}^2}
      | c = Integration by Substitution
}}
{{eqn | r = 4 a \int \frac {\d u} {\sqrt u \paren {u + 4 a c - b^2}^2}
      | c = simplifying
}}
{{end-eqn}}
Recall the result Primitive of $\dfrac 1 {\paren {p x + q}^n \sqrt {a x + b} }$:
:$\ds \int \frac {\d x} {\paren {p x + q}^n \sqrt {a x + b} } = \frac {\sqrt {a x + b} } {\paren {n - 1} \paren {a q - b p} \paren {p x + q}^{n - 1} } + \frac {\paren {2 n - 3} a} {2 \paren {n - 1} \paren {a q - b p} } \int \frac {\d x} {\paren {p x + q}^{n - 1} \sqrt {a x + b} }$
Let:
{{begin-eqn}}
{{eqn | l = x
      | o = =:
      | r = u
}}
{{eqn | l = a
      | o = =:
      | r = 1
}}
{{eqn | l = b
      | o = =:
      | r = 0
}}
{{eqn | l = p
      | o = =:
      | r = 1
}}
{{eqn | l = q
      | o = =:
      | r = 4 a c - b^2
}}
{{eqn | l = n
      | o = =:
      | r = 2
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = 4 a \int \frac {\d u} {\sqrt u \paren {u + 4 a c - b^2}^2}
      | r = 4 a \paren {\frac {\sqrt u} {\paren {4 a c - b^2} \paren {u + 4 a c - b^2} } + \frac 1 {2 \paren {4 a c - b^2} } \int \frac {\d u} {\paren {u + 4 a c - b^2} \sqrt u} }
      | c = 
}}
{{eqn | r = 4 a \paren {\frac z {\paren {4 a c - b^2} \paren {z^2 + 4 a c - b^2} } + \frac 1 {2 \paren {4 a c - b^2} } \int \frac {2 z \rd z} {\paren {z^2 + 4 a c - b^2} z} }
      | c = substituting for $z$
}}
{{eqn | r = 4 a \frac {2 a x + b} {\paren {4 a c - b^2} \paren {\paren {2 a x + b}^2 + 4 a c - b^2} } + \frac {2 a} {4 a c - b^2} \int \frac {4 a \rd x} {\paren {\paren {2 a x + b}^2 + 4 a c - b^2} }
      | c = substituting for $x$
}}
{{eqn | r = \frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {\paren {a x^2 + b x + c} }
      | c = Completing the Square
}}
{{end-eqn}}
{{qed}}
Theorem: Completing the Square

Let $a, b, c, x$ be real numbers with $a \ne 0$.
Then:
:$a x^2 + b x + c = \dfrac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}$
This process is known as '''completing the square'''.

Proof:
{{begin-eqn}}
{{eqn | l = a x^2 + b x + c
      | r = \frac {4 a^2 x^2 + 4 a b x + 4 a c} {4 a}
      | c = multiplying top and bottom by $4 a$
}}
{{eqn | r = \frac {4 a^2 x^2 + 4 a b x + b^2 + 4 a c - b^2} {4 a}
      | c = 
}}
{{eqn | r = \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x over square of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} = \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}$

Proof:
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a x^2 + b x + c
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = 2 a x + b
      | c = Derivative of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {2 a x \rd x} {\paren {a x^2 + b x + c}^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {\paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd z} {\paren {2 a x + b} z^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = Integration by Substitution from $(1)$
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\d z} {z^2} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = simplification
}}
{{eqn | r = \frac {-1} {2 a z} - \frac b {2 a} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a z} - \frac b {2 a} \paren {\frac {2 a x + b} {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } + \frac {2 a} {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c} }
      | c = Primitive of $\dfrac 1 {\paren {a x^2 + b x + c}^2}$
}}
{{eqn | r = \frac {-1} {2 a \paren {a x^2 + b x + c} } - \frac {b \paren {2 a x + b} } {2 a \paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = substituting for $z$ and simplifying
}}
{{eqn | r = \frac {b^2 - 4 a c - 2 a b x - b^2} {2 a \paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = common denominator
}}
{{eqn | r = \frac {-\paren {b x + 2 c} } {\paren {4 a c - b^2} \paren {a x^2 + b x + c} } - \frac b {4 a c - b^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Odd Power of x over Power of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {x^{2 n - 1} \rd x} {\paren {a x^2 + b x + c}^n} = \frac 1 a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^{n - 1} } - \frac c a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^n} - \frac b a \int \frac {x^{2 n - 2} \rd x} {\paren {a x^2 + b x + c}^n}$

Proof:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {x^{2 n - 1} \rd x} {\paren {a x^2 + b x + c}^n}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {x^{2 n - 3} a x^2 \rd x} {\paren {a x^2 + b x + c}^n}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 a \int \frac {x^{2 n - 3} \paren {a x^2 + b x + c - b x - c} \rd x} {\paren {a x^2 + b x + c}^n}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^{n - 1} } - \frac c a \int \frac {x^{2 n - 3} \rd x} {\paren {a x^2 + b x + c}^n} - \frac b a \int \frac {x^{2 n - 2} \rd x} {\paren {a x^2 + b x + c}^n}
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x by square of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} = \frac 1 {2 c \paren {a x^2 + b x + c} } - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }$

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}
      | r = \int \frac {c \rd x} {c x \paren {a x^2 + b x + c}^2}
      | c = multiplying top and bottom by $c$
}}
{{eqn | r = \frac 1 c \int \frac {c \rd x} {x \paren {a x^2 + b x + c}^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 c \int \frac {a x^2 + b x + c - a x^2 - b x} {x \paren {a x^2 + b x + c}^2} \rd x
      | c = adding and subtracting $a x^2 + b x$
}}
{{eqn | r = \frac 1 c \int \frac {\paren {a x^2 + b x + c} \rd x} {x \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x \paren {a x^2 + b x + c}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \int \frac {x \rd x} {x \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} } - \frac a c \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2}
      | c = simplification
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = -\frac {2 a} {2 c} \int \frac {x \rd x} {\paren {a x^2 + b x + c}^2} - \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = splitting up the $\dfrac b c$ term
}}
{{eqn | o = 
      | ro= -
      | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = -\frac 1 {2 c} \int \frac {\paren {2 a x + b} \rd x} {\paren {a x^2 + b x + c}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | o = 
      | ro= -
      | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = -\frac 1 {2 c} \paren {\frac {-1} {a x^2 + b x + c} }
      | c = Primitive of Function under its Derivative
}}
{{eqn | o = 
      | ro= -
      | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \frac 1 {2 c \paren {a x^2 + b x + c} }
      | c = simplifying
}}
{{eqn | o = 
      | ro= -
      | r = \frac b {2 c} \int \frac {\d x} {\paren {a x^2 + b x + c}^2} + \frac 1 c \int \frac {\d x} {x \paren {a x^2 + b x + c} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of x squared by square of a x squared plus b x plus c

Let $a \in \R_{\ne 0}$.
Then:
:$\ds \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2} = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}$

Proof:
From Primitive of Reciprocal of Power of x by Power of a x squared plus b x plus c:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^m \paren {a x^2 + b x + c}^n}
      | r = \frac {-1} {\paren {m - 1} c x^{m - 1} \paren {a x^2 + b x + c}^{n - 1} }
}}
{{eqn | o =
      | ro= -
      | r = \frac {\paren {m + 2 n - 3} a} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 2} \paren {a x^2 + b x + c}^n}
}}
{{eqn | o =
      | ro= -
      | r = \frac {\paren {m - n + 2} b} {\paren {m - 1} c} \int \frac {\d x} {x^{m - 1} \paren {a x^2 + b x + c}^n}
}}
{{end-eqn}}
Setting $m = n = 2$:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}
      | r = \frac {-1} {\paren {2 - 1} c x^{2 - 1} \paren {a x^2 + b x + c}^{2 - 1} }
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac {\paren {2 + 2 \times 2 - 3} a} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 2} \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | o = 
      | ro= -
      | r = \frac {\paren {2 - 2 + 2} b} {\paren {2 - 1} c} \int \frac {\d x} {x^{2 - 1} \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \frac {-1} {c x \paren {a x^2 + b x + c} } - \frac {3 a} c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac {2 b} c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Ring of Sets is Closed under Finite Union

Let $\RR$ be a ring of sets.
Let $A_1, A_2, \ldots, A_n \in \RR$.
Then:
:$\ds \bigcup_{j \mathop = 1}^n A_j \in \RR$

Proof:
Proof by induction:
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\ds \bigcup_{j \mathop = 1}^n A_j \in \RR$
$\map P 1$ is true, as this just says $A_1 \in \RR$.
Theorem: Sigma-Ring contains Limit Superior of Sequence of Sets

Let $\RR$ be a $\sigma$-ring.
Let $\sequence {A_n}_{n \mathop \in \N} \in \RR$ be a sequence of sets in $\RR$.
Then:
:$\ds \limsup_{n \mathop \to \infty} A_n \in \RR$

Proof:
Define the sequence of sets:
:$B_i := \ds \bigcup_{n \mathop = i}^\infty A_n$
From $\sigma$-ring axiom $(\text {SR} 3)$: Closure under Countable Unions:
:$\forall i: B_i \in \RR$
From Sigma-Ring is Closed under Countable Intersections:
:$\forall n: \ds \bigcap_{i \mathop = 0}^\infty B_i \in \RR$
Thus:
{{begin-eqn}}
{{eqn | l = \limsup_{n \mathop \to \infty} A_n
      | r = \bigcap_{i \mathop = 0}^\infty \bigcup_{n \mathop = i}^\infty A_n
      | c = 
}}
{{eqn | r = \bigcap_{i \mathop = 0}^\infty B_i
      | c = 
}}
{{eqn | o = \in
      | r = \RR
      | c = 
}}
{{end-eqn}}
{{qed}}
{{explain|precise justification for final steps}}
Theorem: Gamma Function for Non-Negative Integer Argument

The Gamma function satisfies:
:$\map \Gamma z = \dfrac {\map \Gamma {z + 1} } z$
for any $z$ which is not a nonpositive integer.

Proof:
From Gamma Difference Equation:
:$\map \Gamma {z + 1} = z \, \map \Gamma z$
which is valid for all $z \notin \Z_{\le 0}$.
The result follows by dividing by $z$.
Theorem: Gamma Function of Positive Half-Integer

{{begin-eqn}}
{{eqn | l = \map \Gamma {m + \frac 1 2}
      | r = \frac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi
      | c = 
}}
{{eqn | r = \frac {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } {2^m} \sqrt \pi
      | c = 
}}
{{end-eqn}}
where:
:$m + \dfrac 1 2$ is a half-integer such that $m > 0$
:$\Gamma$ denotes the Gamma function.

Proof:
Proof by induction:
For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
:$\map \Gamma {m + \dfrac 1 2} = \dfrac {\paren {2 m}!} {2^{2 m} m!} \sqrt \pi$
Theorem: Gamma Function of Negative Half-Integer

{{begin-eqn}}
{{eqn | l = \map \Gamma {-m + \frac 1 2}
      | r = \frac {\paren {-1}^m 2^{2 m} m!} {\paren {2 m}!} \sqrt \pi
      | c = 
}}
{{eqn | r = \frac {\paren {-1}^m 2^m} {1 \times 3 \times 5 \times \cdots \times \paren {2 m - 1} } \sqrt \pi
      | c = 
}}
{{end-eqn}}
where:
:$-m + \dfrac 1 2$ is a half-integer such that $m > 0$
:$\Gamma$ denotes the Gamma function.

Proof:
Proof by induction:
For all $m \in \Z_{> 0}$, let $\map P m$ be the proposition:
:$\map \Gamma {-m + \dfrac 1 2} = \dfrac {\paren {-1}^m 2^{2 m} m!} {\paren {2 m}!} \sqrt \pi$
Theorem: Reciprocal times Derivative of Gamma Function

Let $z \in \C \setminus \Z_{\le 0}$.
Then:
:$\ds \dfrac {\map {\Gamma'} z} {\map \Gamma z} = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }$
where:
:$\map \Gamma z$ denotes the Gamma function
:$\map {\Gamma'} z$ denotes the derivative of the Gamma function
:$\gamma$ denotes the Euler-Mascheroni constant.

Proof:
{{begin-eqn}}
{{eqn | l = \frac 1 {\map \Gamma z}
      | r = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }
      | c = Weierstrass Form of Gamma Function
}}
{{eqn | ll= \leadsto
      | l = \map \Gamma z
      | r = \frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n}
      | c = reciprocal of both sides
}}
{{eqn | ll= \leadsto
      | l = \map {\Gamma'} z
      | r = \map {\dfrac \d {\d z} } {\frac {e^{-\gamma z} } z \prod_{n \mathop = 1}^\infty \frac {e^{z/n} } {1 + \frac z n} }
      | c = differentiating {{WRT|Differentiation}} $z$
}}
{{eqn | ll= \leadsto
      | l = \map {\Gamma'} z
      | r = -\frac {e^{-\gamma z} \paren {1 + \gamma z} } {z^2} \prod_{n \mathop = 1}^\infty \paren {\frac {e^{z / n} } {\paren {1 + \frac z n} } } + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \paren {\frac z {n \paren {z + n} } \prod_{i \mathop = 1}^\infty \frac {e^{z / i} } {1 + \frac z i} }
      | c = Product Rule for Derivatives
}}
{{eqn | r = -\frac {e^{-\gamma z} \paren {1 + \gamma z} } {z^2} \frac z {e^{-\gamma z} } \map \Gamma z + \frac {e^{-\gamma z} } z \sum_{n \mathop = 1}^\infty \paren {\frac z {n \paren {z + n} } \frac z {e^{-\gamma z} } \map \Gamma z}
      | c = simplifying the continued product
}}
{{eqn | r = -\frac {1 + \gamma z} z \map \Gamma z + \sum_{n \mathop = 1}^\infty \paren {\frac {z \map \Gamma z} {n \paren {z + n} } }
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \frac {\map {\Gamma'} z} {\map \Gamma z}
      | r = -\frac {1 + \gamma z} z + \sum_{n \mathop = 1}^\infty \paren {\frac z {n \paren {z + n} } }
      | c = dividing both sides by $\map \Gamma z$
}}
{{eqn | r = -\gamma - \frac 1 z + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n} }
}}
{{eqn | r = -\gamma + \sum_{n \mathop = 1}^\infty \paren {\frac 1 n - \frac 1 {z + n - 1} }
      | c = rearranging the series
}}
{{end-eqn}}
{{qed}}