第2个回答 2011-02-20
令u=x^2+2x+2,u=(x+1)^2+1,du/dx=2(x+1),dx=du/2(x+1)
原式=∫du/[2(x+1)^2√u]
=∫du/[2(u-1)√u]
=∫d(√u)/(u-1)
令v=√u,则u=v^2
原式=∫dv/(v^2-1)
=[∫dv/(v-1)-∫dv/(v+1)]/2
=[ln(v-1)-ln(v+1)]/2+C
=[ln(√u-1)-ln(√u+1)]/2+C
={ln[√(x^2+2x+2)-1]-ln[√(x^2+2x+2)+1]}/2+C本回答被提问者采纳