求下列不定积分：
(1)∫dx/[2x+√(x+1)]
(2)∫√(ex-1)dx．
(1)令√(x+1)=t，则x=t2-1，dx=2tdt， ∫dx/[2+√(x+1)] =∫[2t/(2+t)]dt =2∫[(t+2-2)/(2+t)]dt =2∫[1-2/(2+t)]dt =2[t-2ln|2+t|]+C =2√(x+1)-4ln[2+√(x+1)]+C． (2)令√(ex-1)=t，则x=ln(t2+1)，dx=2t/(t2+1)dt ∫√(ex-1)dx =∫[t•2t/(t2+1)]dt =2∫[(t2+1-1)/(t2+1)]dt =2∫[1-1/(t2+1)dt =2(t-arctant)+C =2[√(ex-1)-arctan√(ex-1)]+C．
