∫(dx)/(y²-3y-4)=?

1/(y^2-3y-4)=1/(y-4)(y+1)
=a/(y-4)+b/(y+1)
=(ay+a+by-4b)/(y-4)(y+1)
=[(a+b)y+(a-4b)]/(y-4)(y+1)
所以a+b=0,a-4b=1
b=-1/5,a=1/5
所以1/(y^2-3y-4)
=(1/5)[1/(y-4)-1/(y+1)]
所以原式
=(1/5)[∫1/(y-4)-1/(y+1)]dy
=(1/5)[∫1/(y-4)dy-∫1/(y+1)dy]
=(1/5)[∫1/(y-4)d(y-4)-∫1/(y+1)d(y+1)]
=(1/5)ln|y-4|-(1/5)ln|y+1|+C
=(1/5)ln|(y-4)/(y+1)|+C

因为1/(y²-3y-4)=(1/5)[1/(y-4)-1/(y+1)]
于是原式
=(1/5)∫1/(y-4)-1/(y+1)dy
=(1/5)∫1/(y-4)dy-(1/5)∫1/(y+1)dy
=(1/5)ln|y-4|-(1/5)ln|y+1|+C
=(1/5)ln|(y-4)/(y+1)|+C