Primitive of Reciprocal of a x squared plus b x plus c/c equal to 0
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Theorem
Let $a, b \in \R_{\ne 0}$.
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 1
First:
\(\ds c\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {\d x} {a x^2 + b x}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \int \frac {\d x} {x \paren {a x + b} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 b \ln \size {\frac x {a x + b} } + C\) | Primitive of $\dfrac 1 {x \paren {a x + b} }$ |
$\blacksquare$
Proof 2
Let $c = 0$.
From Primitive of $\dfrac 1 {a x^2 + b x + c}$, we have:
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \begin {cases}
\dfrac 2 {\sqrt {4 a c - b^2} } \map \arctan {\dfrac {2 a x + b} {\sqrt {4 a c - b^2} } } + C & : b^2 - 4 a c < 0 \\ \dfrac 1 {\sqrt {b^2 - 4 a c} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 4 a c} } {2 a x + b + \sqrt {b^2 - 4 a c} } } + C & : b^2 - 4 a c > 0 \\ \dfrac {-2} {2 a x + b} + C & : b^2 = 4 a c \end{cases}$
As $b \ne 0$ it follows that $b^2 - 4 a c = b^2 > 0$.
Thus:
\(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \frac 1 {\sqrt {b^2 - 0} } \ln \size {\dfrac {2 a x + b - \sqrt {b^2 - 0} } {2 a x + b + \sqrt {b^2 - 0} } } + C\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 {\sqrt {b^2} } \ln \size {\frac {2 a x + b - b} {2 a x + b + b} } + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 b \ln \size {\frac {a x} {a x + b} } + C\) | further simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 b \paren {\ln \size {\frac x {a x + b} } + \ln \size a} + C\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 b \ln \size {\frac x {a x + b} } + C\) | subsuming $\dfrac 1 b \ln \size a$ into the arbitrary constant |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x^2 + bx + c$: $14.265$