简单不定积分一道

来源:百度知道 编辑:UC知道 时间:2024/05/29 01:10:11
∫[x+(1-x^2)^(1/2)]^(-1)

就这样。

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

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

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)