∫[1/(y^2-3y-4)]dy
=∫{1/[(y-4)(y+1)]}dy
=(1/5)∫{[(y+1)-(y-4)]/[(y-4)(y+1)]}dy
=(1/5)∫[1/(y-4)]dy-(1/5)∫[1/(y+1)]dy
=(1/5)∫[1/(y-4)]d(y-4)-(1/5)∫[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。
追问那1/5是怎么来的,是不是这样的,(y+1-y+4-5)/(y-4)(y+1)dy=1/(y-4)-1/(y+1)-5/(y-4)(y+1)
那么5/(y-4)(y+1)怎么消掉成1/5的,或是其它的解法,我想知道这个过程,谢谢
追答(1/5)[(y+1)-(y-4)]=1。