Primitive of Reciprocal of x cubed by x fourth plus a fourth

Theorem

$\ds \int \frac {\d x} {x^3 \paren {x^4 + a^4} } = \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}$

$\ds \int \frac {\d x} {x^3 \paren {x^4 + a^4} }$

$=$

$\ds \int \frac {a^4 \rd x} {a^4 x^3 \paren {x^4 + a^4} }$

multiplying top and bottom by $a^4$

$\ds $

$=$

$\ds \int \frac {\paren {x^4 + a^4 - x^4} \rd x} {a^4 x^3 \paren {x^4 + a^4} }$

$\ds $

$=$

$\ds \frac 1 {a^4} \int \frac {\paren {x^4 + a^4} \rd x} {x^3 \paren {x^4 + a^4} } - \frac 1 {a^4} \int \frac {x^4 \rd x} {x^3 \paren {x^4 + a^4} }$

$\ds $

$=$

$\ds \frac 1 {a^4} \int \frac {\d x} {x^3} - \frac 1 {a^4} \int \frac {x \rd x} {x^4 + a^4}$

simplifying

$\ds $

$=$

$\ds \frac 1 {a^4} \paren {\frac {-1} {2 x} } - \frac 1 {a^4} \int \frac {x \rd x} {x^4 + a^4}$

$\ds $

$=$

$\ds \frac {-1} {2 a^4 x^2} - \frac 1 {a^4} \paren {\frac 1 {2 a^2} \arctan \frac {x^2} {a^2} }$

$\ds $

$=$

$\ds \frac {-1} {2 a^4 x^2} - \frac 1 {2 a^6} \arctan \frac {x^2} {a^2}$

simplifying

$\blacksquare$