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 $\dfrac a c$

$\blacksquare$


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