∫√(x+1)/(x+2)*dx令√(x+1)=t,则x=t²-1,dx=2tdt原式=2∫t²dt/(t²+1)=2∫[1-1/(t²+1)]dt=2t-2arctant+C=2√(x+1)-2arctan√(x+1)+C
√(x+1)=t∫√(x+1)/(x+2)dx= ∫t/(t²+1)dt²=2∫t²/(t²+1)dt=2∫1-1/(t²+1)dt=2t-2arctant+C
