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...