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

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Theorem

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


Proof

\(\ds \int \frac {x^m \rd x} {\paren {x^n + a^n}^r}\) \(=\) \(\ds \int \frac {x^{m - n} x^n \rd x} {\paren {x^n + a^n}^r}\)
\(\ds \) \(=\) \(\ds \int \frac {x^{m - n} \paren {x^n + a^n - a^n} \rd x} {\paren {x^n + a^n}^r}\)
\(\ds \) \(=\) \(\ds \int \frac {x^{m - n} \paren {x^n + a^n} \rd x} {\paren {x^n + a^n}^r} - a_n \int \frac {x^{m - n} \rd x} {\paren {x^n + a^n}^r}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \int \frac {x^{m - n} \rd x} {\paren {x^n + a^n}^{r - 1} } - a^n \int \frac {x^{m - n} \rd x} {\paren {x^n + a^n}^r}\) simplification

$\blacksquare$


Sources