Primitive of Reciprocal of a x squared plus b x plus c/a equal to 0
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Theorem
- $\ds \int \frac {\d x} {a x^2 + b x + c} = \frac 1 b \ln \size {b x + c} + C$
when $a = 0$.
Proof
| \(\ds a\) | \(=\) | \(\ds 0\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac {\d x} {b x + c}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 b \ln \size {b x + c} + C\) | Primitive of $\dfrac 1 {a x + b}$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x^2 + bx + c$: $14.265$