Primitive of Reciprocal of a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases} \dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end {cases}$
$a$ equal to Zero
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {b x + c} + C$
when $a = 0$.
$b$ equal to Zero
Let $b = 0$.
Then:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases} \dfrac 1 {\sqrt {a c} } \map \arctan {x \sqrt {\dfrac a c} } + C & : a c > 0 \\ \dfrac 1 {2 \sqrt {-a c} } \ln \size {\dfrac {a x - \sqrt {-a c} } {a x + \sqrt {-a c} } } + C & : a c < 0 \\ \dfrac {-1} {a x} + C & : c = 0 \end {cases}$
$c$ equal to Zero
Let $c = 0$.
Then:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {\frac x {a x + b} } + C$
Proof
First:
\(\ds a x^2 + b x + c\) | \(=\) | \(\ds \frac {\paren {2 a x + b}^2 + 4 a c - b^2} {4 a}\) | 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 |
Negative Discriminant
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 $q$ |
$\Box$
Positive Discriminant
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 $q$ |
$\Box$
Zero Discriminant
Let $b^2 - 4 a c = 0$.
Then:
\(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {4 a \rd x} {\paren {2 a x + b}^2}\) | from $(1)$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {-4 a} {2 a \paren {2 a x + b} } + C\) | Primitive of $\dfrac 1 {\paren {a x + b}^2}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {-2} {2 a x + b} + C\) | simplifying |
$\blacksquare$
Also presented as
In some older works, this result can also be seen presented as:
- $\ds \int \frac {\d x} {a x^2 + 2 b x + c}$
where the solution is then developed via the form:
- $\ds \dfrac 1 a \int \frac {\d x} {\paren {x + \frac b a}^2 + \paren {\frac c a - \frac {b^2} {a^2} } }$
Examples
Primitive of $\dfrac 1 {3 x^2 + 4 x + 2}$
- $\ds \int \frac {\d x} {3 x^2 + 4 x + 2} = \dfrac 1 {\sqrt 2} \map \arctan {\dfrac {3 x + 2} {\sqrt 2} } + C$
Primitive of $\dfrac 1 {x^2 + 4 x + 5}$
- $\ds \int \dfrac {\d x} {x^2 + 4 x + 5} = \map \arctan {x + 2} + C$
Primitive of $\dfrac 1 {x^2 + 2 a x + b}$
- $\ds \int \frac {\d x} {x^2 + 2 a x + b} = \dfrac 1 {\sqrt {b - a^2} } \map \arctan {\dfrac {x + a} {\sqrt {b - a^2} } } + C$
where $b > a^2$.
Also see
- Primitive of $\dfrac 1 {a x + b}$ for the case where $a = 0$
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
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x^2 + bx + c$: $14.265$