Question

∫√1+x/1-xdx

Answer 1

令y=√(1+x),则x=y^2-1

原式=∫y/(2-y^2)d(y^2-1)

=2∫y^2/(2-y^2)dy

=-2∫[1+2/(y^2-2)]dy

=-2∫dy-√2∫[1/(y-√2)-1/(y+√2)]dy

=-2y-√2[ln(y-√2)-ln(y+√2)]+C

=-2y-√2ln[(y-√2)/(y+√2)]+C

=-2√(1+x)-√2ln[(√(1+x)-√2)/(√(1+x)+√2)]+C