avewright/proofwiki-math huggingface dataset:
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
We have:
:$\dfrac \d {\d x} \paren {q \sin a x + p \cos a x} = a q \cos a x - a p \sin a x$
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \tan a x}
      | r = \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }
      | c = Tangent is Sine divided by Cosine
}}
{{eqn | r = \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 + q^2} \int \frac {\paren {p^2 + q^2} \cos a x \rd x} {p \cos a x + q \sin a x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 + q^2} \int \frac {p^2 \cos a x + p q \sin a x - p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {p^2 + q^2} \paren {\int \frac {p^2 \cos a x + p q \sin a x} {p \cos a x + q \sin a x} \rd x + \int \frac {-p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x}
      | c = 
}}
{{eqn | r = \frac 1 {p^2 + q^2} \paren {\int p \rd x + \frac q a \int \frac {\map \d {p \cos a x + q \sin a x} } {p \cos a x + q \sin a x} }
      | c = 
}}
{{eqn | r = \frac {p x} {p^2 + q^2} + \frac q {a \paren {p^2 + q^2} } \ln \size {p \cos a x + q \sin a x} + C
      | c = Primitive of Constant and Primitive of Reciprocal
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
Let $u = p^2 + q^2$ and $v = q^2 - p^2$.
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = u + v
      | r = 2 q^2
}}
{{eqn | n = 2
      | l = u - v
      | r = 2 p ^2
}}
{{end-eqn}}
Also:
{{begin-eqn}}
{{eqn | l = u^2 - v^2
      | r = \paren {u + v} \paren {u - v}
}}
{{eqn | l = u^2 - v^2
      | r = \paren {2 q^2} \paren {2 p^2}
      | c = from $\paren 1$ and $\paren 2$
}}
{{eqn | n = 3
      | l = u^2 - v^2
      | r = 4 p^2 q^2
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x}
      | r = \int \frac {\d x} {p^2 \paren {\frac {1 - \cos 2 a x} 2} + q^2 \cos^2 a x}
      | c = Square of Sine
}}
{{eqn | r = \int \frac {\d x} {p^2 \paren {\frac {1 - \cos 2 a x} 2} + q^2 \paren {\frac {1 + \cos 2 a x} 2} }
      | c = Square of Cosine
}}
{{eqn | r = \int \frac {2 \d x} {p^2 - p^2 \cos 2 a x + q^2 + q^2 \cos 2 a x}
      | c = 
}}
{{eqn | r = \int \frac {2 \d x} {p^2 + q^2 + \paren {q^2 - p^2} \cos 2 a x}
      | c = 
}}
{{eqn | r = 2 \paren { \frac 2 {\paren {2 a} \sqrt{u^2 - v^2} } } \map \arctan {\sqrt{ \frac {u - v} {u + v} } \tan \frac {\paren {2 a} x} 2} + C
      | c = Primitive of $\dfrac 1 {p + q \cos a x}$
}}
{{eqn | r = 2 \paren {\frac 1 {a \sqrt{4 p^2 q^2} } } \map \arctan {\sqrt{ \frac {2 p^2} {2 q^2} } \tan a x} + C
      | c = from $\paren 3$
}}
{{eqn | r = \paren {\frac 2 {2 a p q} } \map \arctan {\sqrt{ \frac {p^2} {q^2} } \tan a x} + C
      | c = 
}}
{{eqn | r = \paren {\frac 1 {a p q} } \map \arctan {\frac p q \tan a x} + C
      | c = 
}}
{{eqn | r = \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn	| l = \int \frac {\d x} {p^2 \sin^2 a x + q^2 \cos^2 a x}
	| r = \int \frac {\sec^2 a x \d x} {p^2 \tan^2 a x + q^2}
	| c = multiplying by $\dfrac {\sec^2 a x} {\sec^2 a x}$
}}
{{eqn	| r = \frac 1 a \int \frac {\d t} {p^2 t^2 + q^2}
	| c = substituting $t = \tan a x$
}}
{{eqn	| r = \frac 1 {a p^2} \int \frac {\d t} {t^2 + \paren {\frac q p}^2}
}}
{{eqn	| r = \frac 1 {a p^2} \times \frac p q \map \arctan {\frac {p t} q} + C
	| c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn	| r = \frac 1 {a p q} \map \arctan {\frac {p \tan a x} q} + C
	| c = substituting back for $t$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = x \sqrt {a x^2 + b x + c}
      | r = \frac x {\paren {a x^2 + b x + c}^{-\frac 1 2} }
      | c = 
}}
{{eqn | r = \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {\paren {a x^2 + b x + c}^{-\frac 1 2} \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} }
      | c = 
}}
{{eqn | r = \frac{x \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} }
      | c = 
}}
{{eqn | r = \frac N D
      | c = 
}}
{{eqn-intertext | where:}}
{{eqn | l = N
      | r = \frac {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} x
      | c = 
}}
{{eqn | l = D
      | r = \frac {2 \sqrt c + \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \frac {\d N} {\d x}
      | r = \frac {x \paren {\sqrt c \paren {2 a x + b} \paren {a x^2 + b x + c}^{-\frac 1 2} + b} - \paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x} + b x \sqrt {a x^2 + b x + c} } - \paren {2 \sqrt c \paren {a x^2 + b x + c} + \paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {2 a x^2 + b x - 2 a x^2 - 2 b x - 2 c} + b x \sqrt {a x^2 + b x + c} } - \paren {\paren {b x + 2 c} \sqrt {a x^2 + b x + c} } } {x^2}
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-\frac 1 2} \frac {\paren {\sqrt c \paren {-b x - 2 c} - \paren {\paren {2 c} \sqrt {a x^2 + b x + c} } } } {x^2}
      | c = 
}}
{{eqn | r = -\frac {2 c + \sqrt c \paren {b x + 2 c} \paren {a x^2 + b x + c}^{-\frac 1 2} } {x^2}
      | c = 
}}
{{eqn | r = \sqrt c \times D
      | c = 
}}
{{end-eqn}}
Now provided $N$ is real and non zero:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \sqrt {a x^2 + b x + c} }
      | r = \int \frac D N \rd x
      | c = 
}}
{{eqn | r = \frac {-1} {\sqrt c} \int \frac {\frac {\d N} {\d x} } N \rd x
      | c = 
}}
{{eqn | r = \frac {-1} {\sqrt c} \ln \size N
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \frac {-1} {\sqrt c} \ln \size {\frac {\paren {2 \sqrt c \sqrt {a x^2 + b x + c} + b x + 2 c} } x}
      | c = 
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {a x + b}^2}
      | r = \int \frac {b \rd x} {b x \paren {a x + b}^2}
      | c = multiplying top and bottom by $b$
}}
{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b}^2}
      | c = adding and subtracting $a x$
}}
{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b}^2} - \frac a b \int \frac {x \rd x} {x \paren {a x + b}^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 b \int \frac {\d x} {x \paren {a x + b} } - \frac a b \int \frac {\d x} {\paren {a x + b}^2}
      | c = simplifying
}}
{{eqn | r = \frac 1 b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } - \frac a b \int \frac {\d x} {\paren {a x + b}^2} + C
      | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$
}}
{{eqn | r = \frac 1 {b^2} \ln \size {\frac x {a x + b} } - \frac a b \paren {-\frac 1 {a \paren {a x + b} } } + C
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | r = \frac 1 {b \paren {a x + b} } + \frac 1 {b^2} \ln \size {\frac x {a x + b} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {a x + b} }
      | r = \int \frac {b \rd x} {b x \paren {a x + b} }
      | c = multiplying top and bottom by $b$
}}
{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x \paren {a x + b} }
      | c = adding and subtracting $a x$
}}
{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x \paren {a x + b} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 b \int \frac {\d x} x - \frac a b \int \frac {\d x} {a x + b}
      | c = simplifying
}}
{{eqn | r = \frac 1 b \ln \size x - \frac a b \int \frac {\d x} {a x + b} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 b \ln \size x - \frac 1 b \ln \size {a x + b} + C
      | c = Primitive of $\dfrac 1 {a x + b}$
}}
{{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
      | r = \int \paren {\frac 1 {a^2 x} - \frac x {a^2 \paren {x^2 + a^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} {x^2 + a^2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^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 {x^2 + a^2} } + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
      | r = \int \frac {a^2 \rd x} {a^2 x \paren {x^2 + a^2} }
      | c = multiplying top and bottom by $a^2$
}}
{{eqn | r = \int \frac {\paren {x^2 + a^2 - x^2} \rd x} {a^2 x \paren {x^2 + a^2} }
      | c = adding and subtracting $x^2$
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\paren {x^2 + a^2} \rd x} {x \paren {x^2 + a^2} } - \frac 1 {a^2} \int \frac {x^2 \rd x} {x \paren {x^2 + a^2} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 {a^2} \int \frac {\d x} x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^2}
      | c = simplifying
}}
{{eqn | r = \frac 1 {a^2} \ln \size x - \frac 1 {a^2} \int \frac {x \rd x} {x^2 + a^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 {x^2 + a^2} } + C
      | c = Primitive of $\dfrac x {x^2 + a^2}$
}}
{{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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
From Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$:
:$\ds \int \frac {\d x} {x \paren {x^n + a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n} {x^n + a^n} } + C$
So:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x \paren {x^2 + a^2} }
      | r = \frac 1 {2 a^2} \ln \size {\frac {x^2} {x^2 + a^2} } + C
      | c = Primitive of $\dfrac 1 {x \paren {x^n + a^n} }$ with $n = 2$
}}
{{eqn | r = \frac 1 {2 a^2} \map \ln {\frac {x^2} {x^2 + a^2} } + C
      | c = Absolute Value of Even Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^3 \paren {a x + b} }
      | r = \int \frac {b \rd x} {b x^3 \paren {a x + b} }
      | c = multiplying top and bottom by $b$
}}
{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x^3 \paren {a x + b} }
      | c = adding and subtracting $a x$
}}
{{eqn | r = \frac 1 b \int \frac {\paren {a x + b} \rd x} {x^3 \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x^3 \paren {a x + b} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 b \int \frac {\d x} {x^3} - \frac a b \int \frac {\d x} {x^2 \paren {a x + b} }
      | c = simplification
}}
{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \int \frac {\d x} {x^2 \paren {a x + b} } + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {2 b x^2} - \frac a b \paren {-\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} } + C
      | c = Primitive of $\dfrac 1 {x^2 \paren {a x + b} }$
}}
{{eqn | r = -\frac 1 {2 b x^2} + \frac a {b^2 x} - \frac {a^2} {b^3} \ln \size {\frac {a x + b} x} + C
      | c = multiplying out
}}
{{eqn | r = \frac a {b^2 x} - \frac 1 {2 b x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
      | c = Difference of Logarithms
}}
{{eqn | r = \frac {2 a x - b} {2 b^2 x^2} + \frac {a^2} {b^3} \ln \size {\frac x {a x + b} } + C
      | c = rearranging
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
Let:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sinh \theta
      | c = Derivative of Hyperbolic Cosine
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = a \cosh \theta
}}
{{eqn | ll= \leadsto
      | l = \sqrt {x^2 - a^2}
      | r = \sqrt {a^2 \paren {\cosh^2 \theta - 1} }
      | c = 
}}
{{eqn | r = \sqrt {a^2 \sinh^2 \theta}
      | c = Difference of Squares of Hyperbolic Cosine and Sine
}}
{{eqn | r = a \sinh \theta
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \sqrt {x^2 - a^2} }
      | r = \int \frac {a \sinh \theta \rd \theta} {a^2 \cosh^2 \theta \cdot a \sinh \theta}
      | c = Integration by Substitution
}}
{{eqn | r = \dfrac 1 {a^2} \int \frac {\rd \theta} {\cosh^2 \theta}
      | c = simplifying
}}
{{eqn | r = \dfrac 1 {a^2} \tanh \theta + C
      | c = Primitive of $\dfrac 1 {\cosh^2 \theta}$
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \sinh \theta} {a \cosh \theta} + C
      | c = {{Defof|Hyperbolic Tangent|index = 2}}
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {\sqrt {x^2 - a^2} } x + C
      | c = substituting for $a \sinh \theta$ and $a \cosh \theta$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

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

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a x + b} }
      | r = \int \frac {b \rd x} {b x^2 \paren {a x + b} }
      | c = multiplying top and bottom by $b$
}}
{{eqn | r = \int \frac {\paren {a x + b - a x} \rd x} {b x^2 \paren {a x + b} }
      | c = adding and subtracting $a x$
}}
{{eqn | r = \frac 1 b \int \frac {\paren {a x + b } \rd x} {x^2 \paren {a x + b} } - \frac a b \int \frac {x \rd x} {x^2 \paren {a x + b} }
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 b \int \frac {\d x} {x^2} - \frac a b \int \frac {\d x} {x \paren {a x + b} }
      | c = simplification
}}
{{eqn | r = -\frac 1 {b x} - \frac a b \int \frac {\d x} {x \paren {a x + b} } + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {b x} - \frac a b \paren {\frac 1 b \ln \size {\frac x {a x + b} } } + C
      | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$
}}
{{eqn | r = -\frac 1 {b x} + \frac a {b^2} \ln \size {\frac {a x + b} x} + C
      | c = Logarithm of Reciprocal and simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

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: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
Let $\size x > a$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \coth^{-1} {\frac x a}
      | c = {{Defof|Real Inverse Hyperbolic Cotangent}}, which is defined where $\size {\dfrac x a} > 1$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = a \coth u
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d u}
      | r = -a \csch^2 u
      | c = Derivative of Hyperbolic Cotangent
}}
{{eqn | ll= \leadsto
      | l = \int \frac 1 {x^2 - a^2} \rd x
      | r = \int \frac {-a \csch^2 u} {a^2 \coth^2 u - a^2} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = -\frac a {a^2} \int \frac {\csch^2 u} {\coth^2 u - 1} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 a \int \frac {\csch^2 u} {\csch^2 u} \rd u
      | c = Difference of Squares of Hyperbolic Cotangent and Cosecant
}}
{{eqn | r = -\frac 1 a \int \rd u
}}
{{eqn | r = -\frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = -\frac 1 a \coth^{-1} {\frac x a} + C
      | c = Definition of $u$
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
Let $x > a$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 - a^2}
      | r = -\frac 1 a \arcoth {\frac x a} + C
      | c = Primitive of Reciprocal of $x^2 - a^2$ in $\arcoth$ form
}}
{{eqn | r = -\frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {x + a} {x - a} } } + C
      | c = $\arcoth {\dfrac x a}$ in Logarithm Form
}}
{{eqn | r = -\frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {2 a} \map \ln {\frac {x - a} {x + a} } + C
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C
      | c = as $\dfrac {x - a} {x + a} > 0$ for $x > a$
}}
{{end-eqn}}
Let $x < -a$.
Let $z = -x$.
Then:
:$\d x = -\d z$
and we then have:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 - a^2}
      | r = \int \frac {-\d z} {\paren {-z}^2 - a^2}
      | c = Integration by Substitution
}}
{{eqn | r = -\int \frac {\d z} {z^2 - a^2}
      | c = simplifying
}}
{{eqn | r = -\frac 1 {2 a} \map \ln {\frac {z - a} {z + a} } + C
      | c = from above
}}
{{eqn | r = \frac 1 {2 a} \map \ln {\frac {z + a} {z - a} } + C
      | c = Logarithm of Reciprocal
}}
{{eqn | r = \frac 1 {2 a} \map \ln {\frac {-x + a} {-x - a} } + C
      | c = substituting $-x$ back for $z$
}}
{{eqn | r = \frac 1 {2 a} \map \ln {\frac {\size {x - a} } {\size {x + a} } } + C
      | c = as $x < -a$
}}
{{eqn | r = \frac 1 {2 a} \ln \size {\frac {x - a} {x + a} } + C
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Primitive of Reciprocal

:$\ds \int \frac {\d x} x = \ln \size x + C$
for $x \ne 0$.

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 - a^2}
      | r = \int \frac {\d x} {\paren {x - a} \paren {x + a} }
      | c = Difference of Two Squares
}}
{{eqn | r = \int \frac {\d x} {2 a \paren {x - a} } - \int \frac {\d x} {2 a \paren {x + a} }
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\d x} {x - a} - \frac 1 {2 a} \int \frac {\d x} {x + a}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {2 a} \ln \size {x - a} - \frac 1 {2 a} \ln \size {x + a} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \dfrac 1 {2 a} \ln \size {\dfrac {x - a} {x + a} } + C
      | c = Difference of Logarithms
}}
{{end-eqn}}
{{qed}}