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

Theorem

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

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

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 14$: Integrals involving $x^2 + a^2$: $14.142$