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