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

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Theorem

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

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}$


Proof 1

First:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {\d x} {a x^2 + c}\)
\(\ds \) \(=\) \(\ds \frac 1 a \int \frac {\d x} {x^2 + \frac c a}\) Primitive of Constant Multiple of Function


Let $a c > 0$.

Then $\dfrac c a > 0$ and:

\(\ds \frac c a\) \(>\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds \frac c a\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {\frac c a}\)


Thus:

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

$\Box$


Let $a c < 0$.

Then $\dfrac c a > 0$ and:

\(\ds \frac c a\) \(<\) \(\ds 0\)
\(\ds \leadsto \ \ \) \(\ds -\frac c a\) \(=\) \(\ds q^2\) for some $q \in \R$
\(\ds \leadsto \ \ \) \(\ds q\) \(=\) \(\ds \sqrt {-\frac c a}\)


Thus:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \frac 1 a \int \frac {\d x} {x^2 - q^2}\)
\(\ds \) \(=\) \(\ds \frac 1 {2 a q} \ln \size {\frac {z - q} {z + q} } + C\) Primitive of $\dfrac 1 {x^2 - a^2}$
\(\ds \) \(=\) \(\ds \frac 1 {2 a \sqrt {-\frac c a} } \ln \size {\frac {z - \sqrt {-\frac c a} } {z + \sqrt {-\frac c a} } } + C\) substituting for $q$
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C\) simplifying

$\Box$


Let $c = 0$.

Then:

\(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {\d x} {a x^2}\)
\(\ds \) \(=\) \(\ds \dfrac {-1} {a x} + C\) Primitive of Power

$\blacksquare$


Proof 2

Let $b = 0$.

From Primitive of Reciprocal of a x squared plus b x plus c, we have:

$\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}$


Let $a c > 0$.

Then $b^2 - 4 a c = 0 - 4 a c < 0$ and so:

\(\ds \int \frac {\d x} {a x^2 + 0 x + c}\) \(=\) \(\ds \frac 2 {\sqrt {4 a c - 0^2} } \map \arctan {\frac {2 a x + 0} {\sqrt {4 a c - 0^2} } } + C\)
\(\ds \) \(=\) \(\ds \frac 2 {\sqrt {4 a c} } \map \arctan {\frac {2 a x} {\sqrt {4 a c} } } + C\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C\) simplifying

$\Box$


Let $a c < 0$.

Then $b^2 - 4 a c = 0 - 4 a c > 0$ and so:

\(\ds \int \frac {\d x} {a x^2 + 0 x + c}\) \(=\) \(\ds \frac 1 {\sqrt {0^2 - 4 a c} } \ln \size {\frac {2 a x + 0 - \sqrt {0^2 - 4 a c} } {2 a x + 0 + \sqrt {0^2 - 4 a c} } } + C\)
\(\ds \) \(=\) \(\ds \frac 1 {\sqrt {- 4 a c} } \ln \size {\frac {2 a x - \sqrt {- 4 a c} } {2 a x + \sqrt {- 4 a c} } } + C\)
\(\ds \) \(=\) \(\ds \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C\) simplifying

$\Box$


Let $c = 0$.

Then $b^2 - 4 a c = 0 - 0 = 0$ and so:

\(\ds \int \frac {\d x} {a x^2 + 0 x + 0}\) \(=\) \(\ds \frac {-2} {2 a x + 0} + C\)
\(\ds \) \(=\) \(\ds \frac {-1} {a x} + C\)

$\blacksquare$


Sources

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