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简单不定积分一道
∫[x+(1-x^2)^(1/2)]^(-1)
就这样.
人气:324 ℃ 时间:2020-03-28 20:47:28
解答
令x=sint,dx=costdt ∫1/[x+√(1-x^2)]dx=∫cost/(sint+cost)dt---A(1) A=∫cost/(sint+cost)dt=∫[(cost+sint)-sint]/(sint+cost)dt=∫1-[sint/(sint+cost)]dt=t-∫sint/(sint+cost)dt (2) A=∫cost/(sint+cost)dt=...
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