Primitive of Power of x over Power of a squared minus x squared

Theorem

$\ds \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n} = a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }$

for $x^2 < a^2$.

Proof

\(\ds \int \frac {x^m \rd x} {\paren {a^2 - x^2}^n}\)

\(=\)

\(\ds \int \frac {x^{m - 2} \paren {x^2} \rd x} {\paren {a^2 - x^2}^n}\)

\(\ds \)

\(=\)

\(\ds \int \frac {x^{m - 2} \paren {a^2 + x^2 - a^2} \rd x} {\paren {a^2 - x^2}^n}\)

\(\ds \)

\(=\)

\(\ds \int \frac {x^{m - 2} \paren {a^2 - \paren {a^2 - x^2} } \rd x} {\paren {a^2 - x^2}^n}\)

\(\ds \)

\(=\)

\(\ds a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \paren {a^2 - x^2} \rd x} {\paren {a^2 - x^2}^n}\)

Linear Combination of Primitives

\(\ds \)

\(=\)

\(\ds a^2 \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^n} - \int \frac {x^{m - 2} \rd x} {\paren {a^2 - x^2}^{n - 1} }\)

simplifying

$\blacksquare$

Also see

• Primitive of $\dfrac {x^m} {\paren {x^2 - a^2}^n}$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $a^2 - x^2$, $x^2 < a^2$: $14.180$