为自己加油

∫₂^+∞1/(x²+x-2)dx=_____．

(2/3)ln2。 由于 ∫[1/(x^{2/sup>+x-2)]dx =(1/3)∫[1/(x-1)]dx-1/3∫[1/(x+2)]dx =(1/3)ln|(x-1)/(x+2)|+C， 所以 ∫₂+∞1/(x2+x-2)dx =lim_b→+∞∫₂b1/(x2+x-2)dx =lim_b→+∞[1/3ln|(x-1)/(x+2)||₂b] =1/3lim_b→+∞[ln (b-1)/(b+2)-ln 1/4] =1/3(0-ln1/4)=2/3ln2．}