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

Theorem

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

for $x^2 > a^2$.

Proof

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

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

Linear Combination of Primitives

\(\ds \)

\(=\)

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

simplifying

$\blacksquare$

Also see

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

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 - a^2$, $x^2 > a^2$: $14.161$