Đặt t=x2 suy ra:
dt=2xdx;x2−3x(x4+3x2+2)dx=(t−3)dtt(t+1)(t+2)
Khi đó:
∫f(x)dx=∫t−3t(t+1)(t+2)dt
Ta có:
t−3t(t+1)(t+2)=at+bt+1+ct+2=(a+b+c)2+(3a+2b+c)t+2at(t+1)(t+2)
Đồng nhất thức ta có:
{a+b+c=03a+2b+c=12a=−3↔{a=−32b=4c=−52
→t−3t(t+1)(t+2)=−321t+4t+1−521t+2
Do đó:
∫f(x)dx=∫(−321t+4t+1−521t+2)dt=−32ln|t|+4ln|t+1|−52ln|t+2|+C
=−32ln(x2)+4ln(x2+1)−52ln(x2+2)+C