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

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 $b^2 \ne 4 a c$.

First Stage of 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