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求不定积分∫(3,2) dx/(x²+x-2)
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∫<2,3> dx/(x²+x-2)
=(1/3)∫<2,3>[1/(x-1)-1/(x+2)]dx
=(1/3)ln|(x-1)/(x+2)|<2,3>
=(1/3)[ln(2/5)-ln(1/4)]
=(1/3)[ln2-ln5+2ln2]
=(1/3)(3ln2-ln5).
=(1/3)∫<2,3>[1/(x-1)-1/(x+2)]dx
=(1/3)ln|(x-1)/(x+2)|<2,3>
=(1/3)[ln(2/5)-ln(1/4)]
=(1/3)[ln2-ln5+2ln2]
=(1/3)(3ln2-ln5).
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