Primitive of Power of x over a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {x^m \rd x} {a x^2 + b x + c} = \frac {x^{m - 1} } {\paren {m - 1} a} - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}$
Proof
\(\ds \int \frac {x^m \rd x} {a x^2 + b x + c}\) | \(=\) | \(\ds \int \frac 1 a \frac {a x^m \rd x} {a x^2 + b x + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {x^{m - 2} a x^2 \rd x} {a x^2 + b x + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c - b x - c} \rd x} {a x^2 + b x + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int \frac {x^{m - 2} \paren {a x^2 + b x + c} \rd x} {a x^2 + b x + c} - \frac b a \int \frac {x^{m - 2} x \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}\) | Linear Combination of Primitives | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 a \int x^{m - 2} \rd x - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}\) | simplification | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{m - 1} } {\paren {m - 1} a} - \frac b a \int \frac {x^{m - 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{m - 2} \rd x} {a x^2 + b x + c}\) | Primitive of Power |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x^2 + b x + c$: $14.268$