avewright/proofwiki-math huggingface dataset:
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} {\sqrt {x^2 - a^2} }
      | r = \cosh^{-1} {\frac x a} + C'
      | c = Primitive of Reciprocal of $\sqrt {x^2 - a^2}$: $\cosh^{-1}$ form
}}
{{eqn | r = \map \ln {\frac x a + \sqrt {\paren {\frac x a}^2 - 1} } + C'
      | c = {{Defof|Real Inverse Hyperbolic Cosine}}
}}
{{eqn | r = \map \ln {\frac x a + \sqrt {\frac {x^2 - a^2} {a^2} } } + C'
      | c = rearrangement
}}
{{eqn | r = \map \ln {\frac {x + \sqrt {x^2 - a^2} } a} + C'
      | c = rearrangement
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } - \ln a + C'
      | c = Difference of Logarithms
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 - a^2} } + C
      | c = putting $C = -\ln a + C'$
}}
{{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 = x
      | r = a \tan \theta
}}
{{eqn | ll= \leadsto
      | l = \frac {\d x} {\d \theta}
      | r = a \sec^2 \theta
      | c = Derivative of Tangent Function
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\paren {\sqrt {x^2 + a^2} }^3}
      | r = \int \frac {a \sec^2 \theta \rd \theta} {\sqrt {a^2 \tan^2 \theta + a^2}^3}
      | c = Integration by Substitution
}}
{{eqn | r = \int \frac {a \sec^2 \theta \rd \theta} {a^3 \sec^3 \theta}
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \dfrac 1 {a^2} \int \cos \theta \rd \theta
      | c = {{Defof|Real Secant Function}} and simplification
}}
{{eqn | r = \dfrac 1 {a^2} \sin \theta + C
      | c = Primitive of $\cos x$
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \sin \theta} {\cos \theta} \dfrac {\cos \theta} a + C
      | c = 
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {a \sec \theta} + C
      | c = Tangent is Sine divided by Cosine, {{Defof|Real Secant Function}}
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \sec^2 \theta} } + C
      | c = 
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac {a \tan \theta} {\sqrt {a^2 \tan^2 \theta + a^2} } + C
      | c = Difference of Squares of Secant and Tangent
}}
{{eqn | r = \dfrac 1 {a^2} \dfrac x {\sqrt {x^2 + a^2} } + C
      | c = substituting $x = a \tan \theta$
}}
{{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} {\sqrt {x^2 + a^2} }
      | r = \arsinh {\frac x a} + C
      | c = Primitive of $\dfrac 1 {\sqrt {x^2 + a^2} }$ in $\arsinh$ form
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } - \ln a + C
      | c = $\arsinh \dfrac x a$ in Logarithm Form
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = subsuming $-\ln a$ into arbitrary constant
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
Let $y^2 = a^2 + x^2$.
Then:
{{begin-eqn}}
{{eqn | l = 2 y \frac {\d y} {\d x}
      | r = 2 x
      | c = Power Rule for Derivatives, Chain Rule for Derivatives
}}
{{eqn | ll= \leadsto
      | l = y \frac {\d y} {\d x}
      | r = x
      | c = simplification
}}
{{eqn | ll= \leadsto
      | l = \frac {\d y} x
      | r = \frac {\d x} y
      | c = 
}}
{{eqn | r = \frac {\d x + \d y} {x + y}
      | c = {{LinkWanted|We don't actually have a result for this}}
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {\sqrt {x^2 + a^2} }
      | r = \int \frac {\d x} y
      | c = substituting for $y$
}}
{{eqn | r = \frac {\d x + \d y} {x + y}
      | c = from above
}}
{{eqn | r = \ln \size {x + y} + C
      | c = Primitive of Function under its Derivative
}}
{{eqn | r = \ln \size {x + \sqrt {x^2 + a^2} } + C
      | c = substituting back
}}
{{eqn | r = \map \ln {x + \sqrt {x^2 + a^2} } + C
      | c = argument of $\ln$ always positive
}}
{{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} {\sin a x + \cos a x}
      | r = \int \frac {\d x} {\sqrt 2 \map \cos {a x - \dfrac \pi 4} }
      | c = Sine of x plus Cosine of x: Cosine Form
}}
{{eqn | r = \frac 1 {\sqrt 2} \int \map \sec {a x - \dfrac \pi 4} \rd x
      | c = Secant is Reciprocal of Cosine
}}
{{end-eqn}}
Let:
{{begin-eqn}}
{{eqn | l = z
      | r = a x - \dfrac \pi 4
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = a
      | c = Derivative of Identity Function <br/>and Derivatives of Function of $a x + b$
}}
{{eqn | ll= \leadsto
      | l = \frac 1 {\sqrt 2} \int \map \sec {a x - \dfrac \pi 4} \rd x
      | r = \frac 1 {a \sqrt 2} \int \sec z \rd z
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\map \tan {\frac z 2 + \frac \pi 4} } + C
      | c = Primitive of $\sec a x$
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\map \tan {\frac 1 2 \paren {a x - \dfrac \pi 4} + \frac \pi 4} } + C
      | c = substituting for $z$
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\map \tan {\frac {a x} 2 + \frac \pi 8} } + 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} {\sin a x + \cos a x}
      | r = \frac 1 a \int \frac {\dfrac {2 \rd u} {1 + u^2} } {\dfrac {2 u} {1 + u^2} + \dfrac {1 - u^2} {1 + u^2} }
      | c = Weierstrass Substitution: $u = \tan \dfrac {a x} 2$
}}
{{eqn | r = \frac 2 a \int \frac {\d u} {- u^2 + 2 u + 1}
      | c = simplifying
}}
{{eqn | r = \frac 2 a \paren {\frac 1 {\sqrt 8} \ln \size {\frac {-2 u + 2 - \sqrt 8} {-2 u + 2 + \sqrt 8} } } + C
      | c = Primitive of $\dfrac 1 {a x^2 + b x + c}$
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\frac {u - 1 + \sqrt 2} {u - 1 - \sqrt 2} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\frac {\tan \dfrac {a x} 2 - \paren {1 - \sqrt 2} } {\tan \dfrac {a x} 2 - \paren {1 + \sqrt 2} } } + C
      | c = substituting for $u$
}}
{{eqn | r = \frac 1 {a \sqrt 2} \ln \size {\frac {\tan \dfrac {a x} 2 - \tan \dfrac \pi 8} {\tan \dfrac {a x} 2 - \tan \dfrac {3 \pi} 8} } + C
      | c = Tangent of $\dfrac \pi 8$ and Tangent of $\dfrac {3 \pi} 8$
}}
{{end-eqn}}
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 $\dfrac x a > 1$
}}
{{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 {\rd x} {a^2 - x^2}
      | r = \int \frac {-a \csch^2 u} {a^2 - a^2 \coth^2 u} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = -\frac a {a^2} \int \frac {\csch^2 u} {-\paren {\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 $\size x < a$.
Let:
{{begin-eqn}}
{{eqn | l = u
      | r = \tanh^{-1} {\frac x a}
      | c = {{Defof|Real Inverse Hyperbolic Tangent}}, which is defined where $\size {\dfrac x a} < 1$
}}
{{eqn | ll=\leadsto
      | l = x
      | r = a \tanh u
      | c = 
}}
{{eqn | ll=\leadsto
      | l = \frac {\d x} {\d u}
      | r = a \sech^2 u
      | c = Derivative of Hyperbolic Tangent
}}
{{eqn | ll=\leadsto
      | l = \int \frac {\rd x} {a^2 - x^2}
      | r = \int \frac {a \sech^2 u} {a^2 - a^2 \tanh^2 u} \rd u
      | c = Integration by Substitution
}}
{{eqn | r = \frac a {a^2} \int \frac {\sech^2 u} {1 - \tanh^2 u} \rd u
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 a \int \frac {\sech^2 u} {\sech^2 u} \rd u
      | c = Sum of Squares of Hyperbolic Secant and Tangent
}}
{{eqn | r = \frac 1 a \int \rd u
}}
{{eqn | r = \frac 1 a u + C
      | c = Integral of Constant
}}
{{eqn | r = \frac 1 a \tanh^{-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 $\size x > a$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a^2 - x^2}
      | r = \frac 1 a \arcoth {\frac x a} + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: $\arcoth$ 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 = \dfrac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + 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:
Let $\size x < a$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a^2 - x^2}
      | r = \frac 1 a \artanh {\frac x a} + C
      | c = Primitive of $\dfrac 1 {a^2 - x^2}$: $\artanh$ form
}}
{{eqn | r = \frac 1 a \paren {\dfrac 1 2 \map \ln {\dfrac {a + x} {a - x} } } + C
      | c = $\artanh \dfrac x a$ in Logarithm Form
}}
{{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - x} } + C
      | c = simplifying
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal

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

Proof:
Let $\size x < a$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a^2 - x^2}
      | r = \int \frac {\d x} {\paren {a + x} \paren {a - x} }
      | c = Difference of Two Squares
}}
{{eqn | r = \int \frac {\d x} {2 a \paren {a + x} } + \int \frac {\d x} {2 a \paren {a - x} }
      | c = Partial Fraction Expansion
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\d x} {a + x} + \frac 1 {2 a} \int \frac {\d x} {a - x}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {2 a} \ln \size {a + x} - \frac 1 {2 a} \ln \size {a - x} + C
      | c = Primitive of Reciprocal
}}
{{eqn | r = \frac 1 {2 a} \map \ln {a + x} - \frac 1 {2 a} \map \ln {a - x} + C
      | c = as both $a + x < 0$ and $a - x < 0$
}}
{{eqn | r = \dfrac 1 {2 a} \map \ln {\dfrac {a + x} {a - 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:
Let $\size x < a$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a^2 - x^2}
      | r = -\int \frac {\d x} {x^2 - a^2}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = -\frac 1 {2 a} \map \ln {\frac {a - x} {a + x} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$ for $\size x < a$
}}
{{eqn | r = \frac 1 {2 a} \map \ln {\frac {x + a} {x - a} } + C
      | c = Logarithm 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 $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\rd x} {\paren {a x + b}^3}
      | r = \frac 1 a \int \frac {\rd u} {u^3}
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 1 a \frac {-1} {2 u^2} + C
      | c = Primitive of Power
}}
{{eqn | r = \frac {-1} {2 a \paren {a x + b}^2} + C
      | c = substituting for $u$
}}
{{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 Power of $a x + b$:
:$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
where $n \ne 1$.
The result follows by setting $n = -3$.
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
Let $u = a x + b$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {\paren {a x + b}^2}
      | r = \frac 1 a \int \frac {\d u} {u^2}
      | c = Primitive of Function of $a x + b$
}}
{{eqn | r = \frac 1 a \frac {-1} u + C
      | c = Primitive of Power
}}
{{eqn | r = -\frac 1 {a \paren {a x + b} } + C
      | c = substituting for $u$
}}
{{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 Power of $a x + b$:
:$\ds \int \paren {a x + b}^n \rd x = \frac {\paren {a x + b}^{n + 1} } {\paren {n + 1} a} + C$
where $n \ne 1$.
The result follows by setting $n = -2$.
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
We aim to use Primitive of $\dfrac 1 {a x^2 + b x + c}$ with:
{{begin-eqn}}
{{eqn | l = a
      | r = 3
}}
{{eqn | l = b
      | r = 4
}}
{{eqn | l = c
      | r = 2
}}
{{end-eqn}}
We note that:
{{begin-eqn}}
{{eqn | l = b^2 - 4 a c
      | r = 4^2 - 4 \times 3 \times 2
}}
{{eqn | r = 16 - 24
}}
{{eqn | r = -8
}}
{{end-eqn}}
Hence from Primitive of $\dfrac 1 {a x^2 + b x + c}$:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
Substituting for $a$, $b$ and $c$ and simplifying:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {3 x^2 + 4 x + 2}
      | r = \dfrac 2 {\sqrt {4 \times 3 \times 2 - 4^2} } \map \arctan {\dfrac {2 \times 3 x + 4} {\sqrt {4 \times 3 \times 2 - 4^2} } } + C
      | c = 
}}
{{eqn | r = \dfrac 2 {\sqrt 8} \map \arctan {\dfrac {6 x + 4} {\sqrt 8} } + C
      | c = 
}}
{{eqn | r = \dfrac 2 {2 \sqrt 2} \map \arctan {\dfrac {2 \paren {3 x + 2} } {2 \sqrt 2} } + C
      | c = 
}}
{{eqn | r = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + 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} {3 x^2 + 4 x + 2}
      | r = \dfrac 1 3 \int \frac {\d x} {x^2 + \frac 4 3 x + \frac 2 3}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac 2 3 - \frac 4 9} }
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \frac 2 9}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \int \frac {\d x} {\paren {x + \frac 2 3}^2 + \paren {\frac {\sqrt 2} 3}^2}
      | c = 
}}
{{eqn | r = \dfrac 1 3 \paren {\dfrac 3 {\sqrt 2} \map \arctan {\dfrac {x + \frac 2 3} {\frac {\sqrt 2} 3} } } + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
First:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2 + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {\d x} {x^2 + \frac c a}
      | c = Primitive of Constant Multiple of Function
}}
{{end-eqn}}
Let $a c > 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 + q^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 {a q} \arctan \frac x q + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 1 {a \sqrt {\frac c a} } \arctan \frac x {\sqrt {\frac c a} } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $a c < 0$.
Then $\dfrac c a > 0$ and:
{{begin-eqn}}
{{eqn | l = \frac c a
      | o = <
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = -\frac c a
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {-\frac c a}
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 a \int \frac {\d x} {x^2 - q^2}
      | c = 
}}
{{eqn | r = \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C
      | c = substituting for $q$
}}
{{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $c = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2}
      | c =
}}
{{eqn | r = \dfrac {-1} {a x} + C
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
Let $b = 0$.
From Primitive of Reciprocal of a x squared plus b x plus c, we have:
:<nowiki>$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end {cases}$
</nowiki>
Let $a c > 0$.
Then $b^2 - 4 a c = 0 - 4 a c < 0$ and so:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + 0 x + c}
      | r = \frac 2 {\sqrt {4 a c - 0^2} } \map \arctan {\frac {2 a x + 0} {\sqrt {4 a c - 0^2} } } + C
      | c = 
}}
{{eqn | r = \frac 2 {\sqrt {4 a c} } \map \arctan {\frac {2 a x} {\sqrt {4 a c} } } + C
      | c = 
}}
{{eqn | r = \frac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $a c < 0$.
Then $b^2 - 4 a c = 0 - 4 a c > 0$ and so:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + 0 x + c}
      | r = \frac 1 {\sqrt {0^2 - 4 a c} } \ln \size {\frac {2 a x + 0 - \sqrt {0^2 - 4 a c} } {2 a x + 0 + \sqrt {0^2 - 4 a c} } } + C
      | c = 
}}
{{eqn | r = \frac 1 {\sqrt {-4 a c} } \ln \size {\frac {2 a x - \sqrt {-4 a c} } {2 a x + \sqrt {-4 a c} } } + C
      | c = 
}}
{{eqn | r = \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Let $c = 0$.
Then $b^2 - 4 a c = 0 - 0 = 0$ and so:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + 0 x + 0}
      | r = \frac {-2} {2 a x + 0} + C
      | c = 
}}
{{eqn | r = \frac {-1} {a x} + C
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
First:
{{begin-eqn}}
{{eqn | l = c
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {\d x} {a x^2 + b x}
      | c = 
}}
{{eqn | r = \int \frac {\d x} {x \paren {a x + b} }
      | c = 
}}
{{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C
      | c = Primitive of $\dfrac 1 {x \paren {a x + b} }$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal

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

Proof:
Let $c = 0$.
From Primitive of $\dfrac 1 {a x^2 + b x + c}$, we have:
:<nowiki>$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \\
\dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \\
\dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c
\end{cases}$</nowiki>
As $b \ne 0$ it follows that $b^2 - 4 a c = b^2 > 0$.
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \frac 1 {\sqrt {b^2 - 0} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 0} } {2 a x + b + \sqrt {b^2 - 0} } } + C
      | c = 
}}
{{eqn | r = \frac 1 {\sqrt {b^2} } \ln \size {\frac {2 a x + b - b} {2 a x + b + b} } + C
      | c = simplifying
}}
{{eqn | r = \frac 1 b \ln \size {\frac {a x} {a x + b} } + C
      | c = further simplifying
}}
{{eqn | r = \frac 1 b \paren {\ln \size {\frac x {a x + b} } + \ln \size a} + C
      | c = simplifying
}}
{{eqn | r = \frac 1 b \ln \size {\frac x {a x + b} } + C
      | c = subsuming $\dfrac 1 b \ln \size a$ into the arbitrary constant
}}
{{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 + q \cos a x}
      | r = \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + \dfrac {p + q} {p - q} }
      | c = Weierstrass Substitution: $u = \tan \dfrac {a x} 2$
}}
{{end-eqn}}
Let $p^2 > q^2$.
Then, by Sign of Quotient of Factors of Difference of Squares:
:$\dfrac {p + q} {p - q} > 0$
Thus, let $\dfrac {p + q} {p - q} = d^2$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \cos a x}
      | r = \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 + d^2}
      | c = 
}}
{{eqn | r = \frac 2 {a \paren {p - q} } \frac 1 d \arctan \frac u d + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 2 {a \paren {p - q} } \frac 1 {\sqrt {\dfrac {p + q} {p - q} } } \map \arctan {\frac {\tan \dfrac {a x} 2} {\sqrt {\dfrac {p + q} {p - q} } } } + C
      | c = substituting for $u$ and $d$
}}
{{eqn | r = \frac 2 {a \sqrt {p^2 - q^2} } \map \arctan {\sqrt {\frac {p - q} {p + q} } \tan \dfrac {a x} 2} + C
      | c = simplifying
}}
{{end-eqn}}
{{qed|lemma}}
Now let $p^2 < q^2$.
Then, by Sign of Quotient of Factors of Difference of Squares:
:$\dfrac {p + q} {p - q} < 0$
Thus, let:
:$-\dfrac {p + q} {p - q} = d^2$
or:
:$\dfrac {q + p} {q - p} = d^2$
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {p + q \cos a x}
      | r = \frac 2 {a \paren {p - q} } \int \frac {\d u} {u^2 - d^2}
      | c = 
}}
{{eqn | r = \frac 2 {a \paren {p - q} } \frac 1 {2 d} \ln \size {\frac {u - d} {u + d} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$
}}
{{eqn | r = \frac 1 {a \paren {p - q} } \frac 1 {\sqrt {\dfrac {q + p} {q - p} } } \ln \size {\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } } + C
      | c = substituting for $u$ and $d$
}}
{{eqn | r = \frac 1 {a \sqrt {q^2 - p^2} } \ln \size {\frac {\tan \dfrac {a x} 2 - \sqrt {\dfrac {q + p} {q - p} } } {\tan \dfrac {a x} 2 + \sqrt {\dfrac {q + p} {q - p} } } } + 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:
First, let $\arctan \dfrac p q = \phi$.
Let $z = a x + \phi$.
{{begin-eqn}}
{{eqn | l = z
      | r = \map \sin {a x + \phi}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d z} {\d x}
      | r = a \map \cos {a x + \phi}
      | c = Derivative of $\sin a x$ etc.
}}
{{eqn | r = a \cos z
      | c = 
}}
{{end-eqn}}
Then:
{{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 = multiplying top and bottom by $\cos a x$
}}
{{eqn | r = \int \frac {\cos a x \rd x} {\sqrt {p^2 + q^2} \map \sin {a x + \phi} }
      | c = Multiple of Sine plus Multiple of Cosine: Sine Form
}}
{{eqn | r = \frac 1 {\sqrt {p^2 + q^2} } \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} }
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {\sqrt {p^2 + q^2} } \paren {\frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C}
      | c = Primitive of $\dfrac {\cos a x} {\map \sin {a x + \phi} }$
}}
{{end-eqn}}
{{finish}}