Primitive of Reciprocal 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:

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


Proof

First:

\(\ds a x^2 + b x + c\) \(=\) \(\ds \dfrac {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} } {4 a}\) Completing the Square
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \int \frac {\d x} {a x^2 + b x + c}\) \(=\) \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2 - \paren {b^2 - 4 a c} }\)


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:

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


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

Then:

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


Thus:

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

$\blacksquare$


Sources

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