Primitive of Reciprocal of Root of a x squared plus b x plus c/Positive Discriminant

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Theorem

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

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

Then:

$\displaystyle \int \frac {\d x} {\sqrt {a x^2 + b x + c} } = \frac 1 {\sqrt a} \map \ln {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C$


Proof

First:

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


Put:

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


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

Thus:

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


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

Then:

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


Thus:

\(\displaystyle \int \frac {\d x} {\sqrt{a x^2 + b x + c} }\) \(=\) \(\displaystyle \frac 1 {\sqrt a} \int \frac {\d z} {\sqrt {z^2 - q^2} }\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt a} \map \ln {z + \sqrt {z^2 - q^2} } + C\) Primitive of $\dfrac 1 {\sqrt {x^2 - a^2} }$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {\paren {2 a x + b}^2 - b^2 - 4 a c} } + C\) substituting for $z$ and $q$
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt a} \map \ln {2 a x + b + \sqrt {4 a \paren {a x^2 + b x + c} } } + C\) Completing the Square
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {\sqrt a} \map \ln {2 \sqrt a \sqrt {a x^2 + b x + c} + 2 a x + b} + C\) rearranging

$\blacksquare$