Question

简单不定积分一道

∫[x+(1-x^2)^(1/2)]^(-1)

就这样。

Answer 1

1/4*log(2*x^2-1)-1/8*2^(1/2)*(-4*(x-1/2*2^(1/2))^2-4*2^(1/2)*(x-1/2*2^(1/2))+2)^(1/2)+1/2*asin(x)+1/4*atanh((1-2^(1/2)*(x-1/2*2^(1/2)))*2^(1/2)/(-4*(x-1/2*2^(1/2))^2-4*2^(1/2)*(x-1/2*2^(1/2))+2)^(1/2))+1/8*2^(1/2)*(-4*(x+1/2*2^(1/2))^2+4*2^(1/2)*(x+1/2*2^(1/2))+2)^(1/2)-1/4*atanh((1+2^(1/2)*(x+1/2*2^(1/2)))*2^(1/2)/(-4*(x+1/2*2^(1/2))^2+4*2^(1/2)*(x+1/2*2^(1/2))+2)^(1/2)
太复杂了，累死了

Answer 2

1/4*log(2*x^2-1)+1/2*asin(x)-1/4*atanh((-2^(1/2)+x)/(1-x^2)^(1/2))-1/4*atanh((2^(1/2)+x)/(1-x^2)^(1/2))

Answer 3

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