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

Theorem

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

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

$=$

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

multiplying top and bottom by $a^n$

$\ds $

$=$

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

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 $

$=$

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

$\blacksquare$