Primitive of Reciprocal of Power of x by Power of Power of x plus Power of a

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Proof

\(\ds \int \frac {\d x} {x^m \paren {x^n + a^n}^r}\) \(=\) \(\ds \int \frac {a^n \rd x} {a^n x^m \paren {x^n + a^n}^r}\) multiplying top and bottom by $a^n$
\(\ds \) \(=\) \(\ds \int \frac {\paren {x^n + a^n - x^n} \rd x} {a^n x^m \paren {x^n + a^n}^r}\) adding and subtracting $x^n$
\(\ds \) \(=\) \(\ds \frac 1 {a^n} \int \frac {\paren {x^n + a^n} \rd x} {x^m \paren {x^n + a^n}^r} - \frac 1 {a^n} \int \frac {x^n \rd x} {x^m \paren {x^n + a^n}^r}\) Linear Combination of Primitives
\(\ds \) \(=\) \(\ds \frac 1 {a^n} \int \frac {\d x} {x^m \paren {x^n + a^n}^{r - 1} } - \frac 1 {a^n} \int \frac {\d x} {x^{m - n} \paren {x^n + a^n}^r}\) simplification

$\blacksquare$


Sources