Question

求不定积分1/(x^2-4)dx=

Answer 1

原式=1/4∫[1/(x-2)-1/(x+2)]dx=1/4[ln(x-2)-ln(x+2)]
=1/4ln[(x-2)/(x+2)]+C

Answer 2

由于x^2-4=(x+2)(x-2)，则1/(x^2-4)=1/4*(x+2)(x-2)，然后积分

追问

过程~·急用···

追答

1/(x^2-4)dx=1/4*(1/(x-2)-1/(x+2))dx=1/4*ln(x-2)-1/4*ln(x+2)