Primitive of Reciprocal of Root of a x squared plus b x plus c/a less than 0

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Theorem

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

Then for $x \in \R$ such that $a x^2 + b x + c > 0$:

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {\size {b^2 - 4 a c} } } } + C$

given that $b^2 \ne 4 a c$.


Proof

Completing the Square

First:

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


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:

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


Positive Discriminant

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

Then:

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


Thus:

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

$\Box$


Negative Discriminant

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

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

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:

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

$\Box$


Zero Discriminant

Suppose that $b^2 - 4 a c = 0$.

Then:

\(\ds a x^2 + b x + c\) \(=\) \(\ds \frac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}\) Completing the Square
\(\ds \) \(=\) \(\ds \frac {\paren {2 a x + b}^2} {4 a}\) as $b^2 - 4 a c = 0$

But we have that:

$\paren {2 a x + b}^2 > 0$

while under our assertion that $a < 0$:

$4 a < 0$

and so:

$a x^2 + b x + c < 0$

Thus on the real numbers $\sqrt {a x^2 + b x + c}$ is not defined.

Hence it follows that:

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

is not defined.

$\Box$


In summary:

For $b^2 - 4 a c > 0: \ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {b^2 - 4 a c} } } + C$
For $b^2 - 4 a c < 0: \ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac {-1} {\sqrt {-a} } \map \arcsin {\frac {2 a x + b} {\sqrt {-\paren {b^2 - 4 a c} } } } + C$

and so by definition of absolute value:

$\ds \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \dfrac {-1} {\sqrt {-a} } \map \arcsin {\dfrac {2 a x + b} {\sqrt {\size {b^2 - 4 a c} } } } + C$

$\blacksquare$


Sources

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