简单不定积分一道∫[x+(1-x^2)^(1/2)]^(-1)就这样.
1个回答
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=∫[(cost-sint)+sint]/(sint+cost)dt=∫(cost-sint)/(sint+cost)dt+∫[sint/(sint+cost)dt=ln(sint+cost)+∫sint/(sint+cost)dt
(1)+(2),得
2A=[t-∫sint/(sint+cost)dt]+[ln(sint+cost)+∫sint/(sint+cost)dt]=t+ln(sint+cost)
则A=[t+ln(sint+cost)]/2={arc sinx+ln[x+√(1-x^2)]}/2
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