Primitive of Power of x over a x squared plus b x plus c

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 $a x^2 + b x + c$: $14.268$