Primitive of Reciprocal of a x squared plus b x plus c

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Theorem

Let $a \in \R_{\ne 0}$.

Then:

$\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}$


$a$ equal to Zero

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

when $a = 0$.


$b$ equal to Zero

Let $b = 0$.

Then:

$\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}$


$c$ equal to Zero

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:

\(\ds a x^2 + b x + c\) \(=\) \(\ds \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}\) Completing the Square
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }\)

Put:

\(\ds z\) \(=\) \(\ds 2 a x + b\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d z} {\d x}\) \(=\) \(\ds 2 a\) Derivative of Power


Let $D = b^2 - 4 a c$.

Thus:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }\) from $(1)$
\(\ds \) \(=\) \(\ds \int \frac {2 \rd z} {z^2 - D}\) Integration by Substitution


Negative Discriminant

Let $b^2 - 4 a c < 0$.

Then:

\(\ds - D\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds - D\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {4 a c - b^2}\) by definition of $D$


Thus:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {2 \rd z} {z^2 + q^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 2 q \arctan \frac z q + C\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C\) substituting for $z$ and $q$


$\Box$


Positive Discriminant

Let $b^2 - 4 a c > 0$.

Then:

\(\ds D\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds D\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {b^2 - 4 a c}\) Definition of $D$


Thus:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {2 \rd z} {z^2 - q^2}\) Integration by Substitution
\(\ds \) \(=\) \(\ds \frac 1 q \ln \size {\frac {z - q} {z + q} } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$: Logarithm Form
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\frac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C\) substituting for $z$ and $q$


$\Box$


Zero Discriminant

Let $b^2 - 4 a c = 0$.

Then:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2}\) from $(1)$
\(\ds \) \(=\) \(\ds \frac {-4 a} {2 a \paren {2 a x + b} } + C\) Primitive of $\dfrac 1 {\paren {a x + b}^2}$
\(\ds \) \(=\) \(\ds \dfrac {-2} {2 a x + b} + C\) simplifying


$\blacksquare$

Also presented as

In some older works, this result can also be seen presented as:

$\ds \int \frac {\d x} {a x^2 + 2 b x + c}$

where the solution is then developed via the form:

$\ds \dfrac 1 a \int \frac {\d x} {\paren {x + \frac b a}^2 + \paren {\frac c a - \frac {b^2} {a^2} } }$


Examples

Primitive of $\dfrac 1 {3 x^2 + 4 x + 2}$

$\ds \int \frac {\d x} {3 x^2 + 4 x + 2} = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C$


Primitive of $\dfrac 1 {x^2 + 4 x + 5}$

$\ds \int \dfrac {\d x} {x^2 + 4 x + 5} = \map \arctan {x + 2} + C$


Primitive of $\dfrac 1 {x^2 + 2 a x + b}$

$\ds \int \frac {\d x} {x^2 + 2 a x + b} = \dfrac 1 {\sqrt {b - a^2} } \map \arctan {\dfrac {x + a} {\sqrt {b - a^2} } } + C$

where $b > a^2$.


Also see


Sources

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