Primitive of x cubed over x squared plus a squared squared/Proof 1

Theorem

$\ds \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2} = \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C$

Let:

$\ds z$

$=$

$\ds x^2 + a^2$

$\ds \leadsto \ \ $

$\ds \frac {\d z} {\d x}$

$=$

$\ds 2 x$

$\ds \leadsto \ \ $

$\ds \int \frac {x^3 \rd x} {\paren {x^2 + a^2}^2}$

$=$

$\ds \int \frac {\paren {z - a^2} } {z^2} \frac {\d z} 2$

$\ds $

$=$

$\ds \frac 1 2 \int \frac {\d z} z - \frac {a^2} 2 \int \frac {\d z} {z^2}$

$\ds $

$=$

$\ds \frac 1 2 \int \frac {\d z} z + \frac {a^2} {2 z} + C$

$\ds $

$=$

$\ds \frac 1 2 \ln z + \frac {a^2} {2 z} + C$

$\ds $

$=$

$\ds \frac {a^2} {2 \paren {x^2 + a^2} } + \frac 1 2 \map \ln {x^2 + a^2} + C$

substituting for $z$

$\blacksquare$