Theorem: Primitive of Reciprocal

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

Proof:
First:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \int \frac {c \rd x} {c x^2 \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^2 \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^2 \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^2 \paren {a x^2 + b x + c}^2} - \frac a c \int \frac {x^2 \rd x} {x^2 \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^2 \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | n = 1
      | r = \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = simplification
}}
{{eqn | o = 
      | ro= -
      | r = \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{end-eqn}}
Next, with a view to obtaining an expression in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
{{begin-eqn}}
{{eqn | l = u
      | r = \frac 1 {\paren {a x^2 + b x + c} }
      | c = 
}}
{{eqn | r = \paren {a x^2 + b x + c}^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {\d u} {\d x}
      | r = -\paren {2 a x + b} \paren {a x^2 + b x + c}^{-2}
      | c = Chain Rule for Derivatives and Derivative of Power
}}
{{eqn | r = \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2}
      | c = simplifying
}}
{{end-eqn}}
and let:
{{begin-eqn}}
{{eqn | l = \frac {\d v} {\d x}
      | r = \frac 1 {x^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = v
      | r = \frac {-1} x
      | c = Primitive of Power
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} }
      | r = \int \frac 1 {\paren {a x^2 + b x + c} } \frac 1 {x^2} \rd x
      | c = 
}}
{{eqn | r = \frac 1 {\paren {a x^2 + b x + c} } \frac {-1} x - \int \frac {-1} x \frac {-\paren {2 a x + b} } {\paren {a x^2 + b x + c}^2} \rd x
      | c = Integration by Parts
}}
{{eqn | n = 2
      | r = \frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2}
      | c = simplification
}}
{{eqn | o = 
      | ro= -
      | r = b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}
      | c = Linear Combination of Primitives
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | o = 
      | r = \int \frac {\d x} {x^2 \paren {a x^2 + b x + c}^2}
      | c = 
}}
{{eqn | r = \frac 1 c \int \frac {\d x} {x^2 \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \paren {a x^2 + b x + c}^2}
      | c = from $(1)$
}}
{{eqn | r = \frac 1 c \paren {\frac {-1} {x \paren {a x^2 + b x + c} } - 2 a \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - b \int \frac {\d x} {x \paren {a x^2 + b x + c}^2} }
      | c = from $(2)$
}}
{{eqn | o = 
      | ro= -
      | r = \frac a c \int \frac {\d x} {\paren {a x^2 + b x + c}^2} - \frac b c \int \frac {\d x} {x \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 = gathering terms
}}
{{end-eqn}}
{{qed}}