答：∫[(x+1)/(x^2+2x)]dx=(1/2) ∫ {1/[(x+1)^2-1]} d [(x+1)^2-1]=(1/2) ln[(x+1)^2-1]+C=(1/2)ln|x^2+2x|+C
∵ ﹙x+1﹚/﹙x²+2x﹚＝﹙1/2﹚﹙x²+2x﹚′/﹙x²+2x﹚∴∫[﹙x+1﹚/﹙x²+2x﹚]dx＝﹙1/2﹚㏑﹙x²+2x﹚+c＝㏑√﹙x²+2x﹚+c
