∫1/[2x+√(1-x^2)]dx的值

Question

∫1/[2x+√(1-x^2)] dx的值

Answer 1

令x=siny,则dx=cosydy
∫1/[2x+√(1-x^2)] dx
=∫1/(2siny+cosy)*cosydy
=∫cosy/(2siny+cosy)dy=A
令∫siny/(2siny+cosy)dy=B
则有A+2B=∫cosy/(2siny+cosy)dy+2∫siny/(2siny+cosy)dy=∫(cosy+2siny)/(2siny+cosy)dy=y+c1
2A-B=∫2cosy/(2siny+cosy)dy-∫siny/(2siny+cosy)dy
=∫(2cosy-siny)/(2siny+cosy)dy=∫(2cosy-siny)/(2siny+cosy)dy
=∫d(2siny+cosy)/(2siny+cosy)=ln|2siny+cosy|+C2
联立解出来A,B,再将y=arcsinx反代即可.