Primitive of Reciprocal of Power of x by a x squared plus b x plus c
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Theorem
Let $a \in \R_{\ne 0}$.
Then:
- $\ds \int \frac {\d x} {x^n \paren {a x^2 + b x + c} } = \frac {-1} {\paren {n - 1} c x^{n - 1} } - \frac b c \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }$
Proof
\(\ds \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }\) | \(=\) | \(\ds \int \frac {x^{-n + 2} \rd x} {a x^2 + b x + c}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x^{-n + 1} } {\paren {-n + 1} a} - \frac b a \int \frac {x^{-n + 1} \rd x} {a x^2 + b x + c} - \frac c a \int \frac {x^{-n} \rd x} {a x^2 + b x + c}\) | Primitive of $\dfrac {x^m} {a x^2 + b x + c}$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \frac c a \int \frac {\d x} {x^n \paren {a x^2 + b x + c} }\) | \(=\) | \(\ds \frac {-1} {\paren {n - 1} a x^{n - 1} } - \frac b a \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }\) | rearranging | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \int \frac {\d x} {x^n \paren {a x^2 + b x + c} }\) | \(=\) | \(\ds \frac {-1} {\paren {n - 1} c x^{n - 1} } - \frac b c \int \frac {\d x} {x^{n - 1} \paren {a x^2 + b x + c} } - \frac a c \int \frac {\d x} {x^{n - 2} \paren {a x^2 + b x + c} }\) | multiplying by $\dfrac a c$ |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a x^2 + b x + c$: $14.271$