Primitive of Reciprocal of a x squared plus b x plus c/Negative Discriminant
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Theorem
Let $a \in \R_{\ne 0}$.
Let $b^2 - 4 a c < 0$.
Then:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C$
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 $q \in \R$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds q\) | \(=\) | \(\ds \sqrt {4 a c - b^2}\) | by 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 q \arctan \frac z q + C\) | Primitive of $\dfrac 1 {x^2 + a^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 {\sqrt {4 a c - b^2} } \map \arctan {\frac {2 a x + b} {\sqrt {4 a c - b^2} } } + C\) | substituting for $z$ and $q$ |
$\blacksquare$
Sources
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.3$ Rules for Differentiation and Integration: Integrals of Rational Algebraic Functions: $3.3.16$