Question

求不定积分∫dx/x-√1-x²

Answer 1

估计原题是：∫dx/[x-√(1-x²)]
设x=sin t 则 dx=cos t dt,
原式= ∫cos t dt/[sin t-cos t]
= -∫[cos^2(t)+sintcost]dt/[cos^2(t) -sin^2(t)]
= -1/2∫[cos(2t)+sin(2t)+1]dt/cos(2t)
=-1/2t+1/4ln|cos(2t)|-1/4ln|sec(2t)+tan(2t)|+C
=-1/2arcsinx+1/4ln|2x^2-1|-1/4ln|1+2x√(1-x^2)|+1/4ln|2x^2-1|+C

因计算量太大，计算结果你再核实一下。