1/(x^2 + x - 2) = 1 / [(x-1)(x+2)] = [1/(x-1) - 1/(x+2)]/3
integral of the above expression is (1/3) * ln [(x-1)/(x+2)]
so the answer is -(1/3) * ln 4
(original problem should not have dx twice)
1/(x^2 + x - 2) = 1 / [(x-1)(x+2)] = [1/(x-1) - 1/(x+2)]/3
integral of the above expression is (1/3) * ln [(x-1)/(x+2)]
so the answer is -(1/3) * ln 4
(original problem should not have dx twice)
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妙,但不用廣義積分法,可能要扣分.
-jinjing-
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12/16/2010 postreply
15:36:59