Question

求不定积分问题

1/(x+1)(x+3)请求它的不定积分

Answer 1

∫1/[(x+1)(x+3)] dx
有理积分法,先化成部分分式
设1/[(x+1)(x+3)]=A/(x+1)+B/(x+3),通分后得
1=A(x+3)+B(x+1)
1=Ax+3A+Bx+B
0x+1=(A+B)x+(3A+B),对比系数得
A+B=0
3A+B=1
解得A=1/2 B=-1/2
∴原式=∫{1/[2(x+3)-1/[2(x+1)]}dx
=(1/2)∫[1/(x+3)]dx-(1/2)∫[1/(x+1)]dx
=(1/2)∫[(1/x+3)]d(x+3)-(1/2)∫[1/(x+1)]d(x+1)
=(1/2)ln|x+3|-(1/2)ln|x+1|+C
=(1/2)ln|(x+3)/(x+1)|+C