Question

不定积分，求大神解一下，这道题？谢谢啦

Answer 1

∫1/(x^2-x-6) dx = ∫ 1/[(x+2)(x-3)] dx = 1/5 * ∫ [1/(x-3) - 1/(x+2)]dx
=1/5 * [ln(x-3) - ln (x+2)] + C
=1/5 * ln [(x-3)/(x+2) ]+ C

追问

你好，能问下1/5是拿来来的鸭

追答

1/[(x+2)(x-3)] =(1/5 * 5)/[(x+2)(x-3)] = 1/5 * [(x+2)-(x-3)]/[(x+2)(x-3)]
=1/5 * [1/(x-3) - 1/(x+2)]