Primitive of Reciprocal of Root of a x squared plus b x plus c/a greater 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} } = \begin {cases}

\dfrac 1 {\sqrt a} \ln \size {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C & : b^2 - 4 a c > 0 \\ \dfrac 1 {\sqrt a} \map \arsinh {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt a} \ln \size {2 a x + b} + C & : b^2 - 4 a c = 0 \end {cases}$ where $\arsinh$ denotes the area hyperbolic sine function.

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

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }\)

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:

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

\(=\)

\(\ds \int \frac {2 \sqrt a \rd x} {\sqrt {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } }\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \int \frac {\d z} {\sqrt a \sqrt {z^2 - D} }\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - D} }\)

Primitive of Constant Multiple of Function