∫(0→4)1/(x∧2-x-2)dx详细步骤

问题描述:

∫(0→4)1/(x∧2-x-2)dx详细步骤

原式=3∫(0→4)[1/(x-2)-1/(x+1)]dx

1/(x^2-x-2)=1/[(x-2)(x+1)}=(1/3)[1/(x-2)-1/(x+1)]=(1/3)[ln(x-2)-ln(x+1)'所以∫(0→4)1/(x∧2-x-2)dx=∫(0→4)d[ln(x-2)/3-ln(x+1)/3]=ln2/3-ln5/3-[ln(-2)/3-ln1/3]好像有点不对?所给x的范围不对...