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

:<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>

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 of a x squared plus b x plus c

:<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>

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 of a x squared plus b x plus c

:<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>

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 of a x squared plus b x plus c/b equal to 0

Let $b = 0$.
Then:
:<nowiki>$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C & : a c > 0 \\ \\
\dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C & : a c < 0 \\ \\
\dfrac {-1} {a x} + C & : c = 0
\end {cases}$
</nowiki>

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 of a x squared plus b x plus c/b equal to 0

Let $b = 0$.
Then:
:<nowiki>$\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C & : a c > 0 \\ \\
\dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C & : a c < 0 \\ \\
\dfrac {-1} {a x} + C & : c = 0
\end {cases}$
</nowiki>

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 of a x squared plus b x plus c/a equal to 0

:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {b x + c} + C$
when $a = 0$.

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

Let $a \in \R_{\ne 0}$.
{{:Primitive of Reciprocal of a x squared plus b x plus c/b equal to 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 x over a x squared plus b x plus c

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

Proof:
First note that by Derivative of Power:
:$(1): \quad \map {\dfrac \d {\d x} } {a x^2 + b x + c} = 2 a x + b$
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {x \rd x} {a x^2 + b x + c}
      | r = \frac 1 {2 a} \int \frac {2 a x \rd x} {a x^2 + b x + c}
      | c = Primitive of Constant Multiple of Function
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b - b} \rd x} {a x^2 + b x + c} 
      | c = 
}}
{{eqn | r = \frac 1 {2 a} \int \frac {\paren {2 a x + b} \rd x} {a x^2 + b x + c} - \frac b {2 a} \int \frac {\d x} {a x^2 + b x + c}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \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}
      | c = Primitive of Function under its Derivative and $(1)$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0

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

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 of a x squared plus b x plus c/c equal to 0

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

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 x squared over a x squared plus b x plus c

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^2 \rd x} {a x^2 + b x + c}
      | r = \int \frac 1 a \paren {1 - \frac {b x + c} {a x^2 + b x + c} } \rd x
      | c = by division
}}
{{eqn | r = \frac 1 a \int \rd x - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac x a - \frac b a \int \frac {x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {\d x} {a x^2 + b x + c}
      | c = Primitive of Constant
}}
{{eqn | r = \frac x a - \frac b a \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 c a \int \frac {\d x} {a x^2 + b x + c}
      | c = Primitive of $\dfrac x {a x^2 + b x + c}$
}}
{{eqn | r = \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} + \frac {b^2} {2 a^2} \int \frac {\d x} {a x^2 + b x + c} - \frac {2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = multiplying out
}}
{{eqn | r = \frac x a - \frac b {2 a^2} \ln \size {a x^2 + b x + c} + \frac {b^2 - 2 a c} {2 a^2} \int \frac {\d x} {a x^2 + b x + c}
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a x squared plus b x plus c

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \dfrac {6 x^2 + 10 x + 5} {3 x^2 + 4 x + 2} \rd x
      | r = \int \dfrac {6 x^2 + 8 x + 4} {3 x^2 + 4 x + 2} \rd x + \int \dfrac {2 x + 1} {3 x^2 + 4 x + 2} \rd x
      | c = Linear Combination of Primitives
}}
{{eqn | r = \int 2 \rd x + \int \dfrac {2 x + 1} {3 x^2 + 4 x + 2} \rd x
      | c = simplification
}}
{{eqn | r = 2 x + \int \dfrac {2 x + 1} {3 x^2 + 4 x + 2} \rd x
      | c = Primitive of Constant
}}
{{eqn | r = 2 x + \dfrac 1 3 \ln \size {3 x^2 + 4 x + 2} - \dfrac 1 {3 \sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C
      | c = Primitive of $\dfrac {2 x + 1} {3 x^2 + 4 x + 2}$
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of x squared over a x squared plus b x plus c

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \dfrac {6 x^2 + 10 x + 5} {3 x^2 + 4 x + 2} \rd x
      | r = 6 \int \dfrac {x^2} {3 x^2 + 4 x + 2} \rd x + 10 \int \dfrac x {3 x^2 + 4 x + 2} \rd x + 5 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = Linear Combination of Primitives
}}
{{eqn | r = 6 \paren {\frac x 3 - \frac 4 {2 \times 3^2} \ln \size {3 x^2 + 4 x + 2} + \frac {4^2 - 2 \times 3 \times 2} {2 \times 3^2} \int \frac {\d x} {3 x^2 + 4 x + 2} } + 10 \int \dfrac x {3 x^2 + 4 x + 2} \rd x + 5 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = Primitive of $\dfrac {x^2} {a x^2 + b x + c}$
}}
{{eqn | r = 2 x - \frac 4 3 \ln \size {3 x^2 + 4 x + 2} + \frac 4 3 \int \frac {\d x} {3 x^2 + 4 x + 2} + 10 \int \dfrac x {3 x^2 + 4 x + 2} \rd x + 5 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = simplification
}}
{{eqn | r = 2 x - \frac 4 3 \ln \size {3 x^2 + 4 x + 2} + 10 \int \dfrac x {3 x^2 + 4 x + 2} \rd x + \dfrac {19} 3 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = further simplification
}}
{{eqn | r = 2 x - \frac 4 3 \ln \size {3 x^2 + 4 x + 2} + 10 \paren {\frac 1 {2 \times 3} \ln \size {3 x^2 + 4 x + 2} - \frac 4 {2 \times 3} \int \frac {\d x} {3 x^2 + 4 x + 2} } + \dfrac {19} 3 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = Primitive of $\dfrac x {a x^2 + b x + c}$
}}
{{eqn | r = 2 x - \frac 4 3 \ln \size {3 x^2 + 4 x + 2} + \frac 5 3 \ln \size {3 x^2 + 4 x + 2} - \frac {20} 3 \int \frac {\d x} {3 x^2 + 4 x + 2} + \dfrac {19} 3 \int \dfrac {\d x} {3 x^2 + 4 x + 2}
      | c = simplification
}}
{{eqn | r = 2 x + \frac 1 3 \ln \size {3 x^2 + 4 x + 2} - \dfrac 1 3 \int \frac {\d x} {3 x^2 + 4 x + 2}
      | c = further simplification
}}
{{eqn | r = 2 x + \frac 1 3 \ln \size {3 x^2 + 4 x + 2} - \dfrac 1 3 \paren {\dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } } + C
      | c = Primitive of $\dfrac 1 {3 x^2 + 4 x + 2}$
}}
{{eqn | r = 2 x + \frac 1 3 \ln \size {3 x^2 + 4 x + 2} - \dfrac 1 {3 \sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C
      | c = simplification
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Power of x over a x squared plus b x plus c

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

Proof:
{{begin-eqn}}
{{eqn | l = \int \frac {x^m \rd x} {a x^2 + b x + c}
      | r = \int \frac 1 a \frac {a x^m \rd x} {a x^2 + b x + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {x^{m - 2} a x^2 \rd x} {a x^2 + b x + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c - b x - c} \rd x} {a x^2 + b x + c}
      | c = 
}}
{{eqn | r = \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c} \rd x} {a x^2 + b x + c} - \frac b a \int \frac {x^{m - 2} x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}
      | c = Linear Combination of Primitives
}}
{{eqn | r = \frac 1 a \int x^{m - 2} \rd x - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}
      | c = simplification
}}
{{eqn | r = \frac {x^{m - 1} } {\paren {m - 1} a} - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}
      | c = Primitive of Power
}}
{{end-eqn}}
{{qed}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant

Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c < 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$

Proof:
Let $b^2 - 4 a c < 0$.
Then:
{{begin-eqn}}
{{eqn | l = - D
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = - D
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {4 a c - b^2}
      | c = by definition of $D$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {2 \rd z} {z^2 + q^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 2 q \arctan \frac z q + C
      | c = Primitive of $\dfrac 1 {x^2 + a^2}$
}}
{{eqn | r = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C
      | c = substituting for $z$ and $q$
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c/Positive Discriminant

Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c > 0$.
Then:
:$\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C$

Proof:
Let $b^2 - 4 a c > 0$.
Then:
{{begin-eqn}}
{{eqn | l = D
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = D
      | r = q^2
      | c = for some $q \in \R$
}}
{{eqn | ll= \leadsto
      | l = q
      | r = \sqrt {b^2 - 4 a c}
      | c = Definition of $D$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {2 \rd z} {z^2 - q^2}
      | c = Integration by Substitution
}}
{{eqn | r = \frac 1 q \ln \size {\frac {z - q} {z + q} } + C
      | c = Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
}}
{{eqn | r = \frac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\frac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C
      | c = substituting for $z$ and $q$
}}
{{end-eqn}}
Theorem: Primitive of Reciprocal of a x squared plus b x plus c/Zero Discriminant

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

Proof:
Let $b^2 - 4 a c = 0$.
Then:
{{begin-eqn}}
{{eqn | l = \int \frac {\d x} {a x^2 + b x + c}
      | r = \int \frac {4 a \rd x} {\paren {2 a x + b}^2}
      | c = from $(1)$
}}
{{eqn | r = \frac {-4 a} {2 a \paren {2 a x + b} } + C
      | c = Primitive of $\dfrac 1 {\paren {a x + b}^2}$
}}
{{eqn | r = \dfrac {-2} {2 a x + b} + C
      | c = simplifying
}}
{{end-eqn}}