Đặt $t=x^4$, suy ra:
$dt=4x^3dx ; \frac{x^3dx}{(x^8-4)^2} =\frac{1}{4}\frac{dt}{(t^2-4)^2} $
Khi đó:
$\int\limits f(x)dx=\frac{1}{4} \int\limits \frac{dt}{(t^2-4)^2} $
Sử dụng đồng nhất thức:
$1=\frac{1}{16}[(t+2)-(t-2)]^2 $
Ta được:
$\int\limits f(x)dx=\frac{1}{64}\int\limits \frac{[(t+2)-(t-2)]^2}{(t^2-4)^2} .dt =\frac{1}{64}\int\limits [\frac{1}{(t-2)^2}-\frac{2}{t^2-4}+ \frac{1}{(t+2)^2}]dt $
$= \frac{1}{64} \int\limits \left[ {\frac{1}{ \left ( t-2 \right )^2 }- \frac{1}{2} \left ( \frac{1}{t-2}-\frac{1}{t+2} \right ) + \frac{1}{ \left ( t+2 \right )^2 } } \right] dt$
$=\frac{1}{64}[-\frac{1}{t-2}-\frac{1}{2}\ln|\frac{t-2}{t+2} |-\frac{1}{t+2}]+C $
$=-\frac{1}{64}(\frac{2t}{t^2-4}+\frac{1}{2}\ln|\frac{t-2}{t+2} |)+C= -\frac{1}{64}(\frac{2x^4}{x^8-4}+\frac{1}{2}\ln|\frac{x^4-2}{x^4+2} |)+C $