Primitive of Reciprocal of x by Power of x minus Power of a

Theorem

$\ds \int \frac {\d x} {x \paren {x^n - a^n} } = \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$

$\ds \int \frac {\d x} {x \paren {x^n - a^n} }$

$=$

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

multiplying top and bottom by $a^n$

$\ds $

$=$

$\ds \int \frac {\paren {-\paren {x^n - a^n} + x^n} \rd x} {a^n x \paren {x^n - a^n} }$

adding and subtracting $x^n$

$\ds $

$=$

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

$\ds $

$=$

$\ds \frac {-1} {a^n} \int \frac {\d x} x + \frac 1 {a^n} \int \frac {x^{n - 1} \rd x} {x^n - a^n}$

simplifying

$\ds $

$=$

$\ds \frac {-1} {a^n} \ln \size x + \frac 1 {a^n} \int \frac {x^{n - 1} \rd x} {x^n - a^n} + C$

$\ds $

$=$

$\ds \frac {-1} {a^n} \ln \size x + \frac 1 {a^n} \paren {\frac 1 n \ln \size {x^n - a^n} } + C$

$\ds $

$=$

$\ds \frac {-1} {n a^n} \ln \size {x^n} + \frac 1 {n a^n} \ln \size {x^n - a^n} + C$

$\ds $

$=$

$\ds \frac 1 {n a^n} \ln \size {\frac {x^n - a^n} {x^n} } + C$

$\blacksquare$