Question

求1/u^2-2u的积分?

Answer 1

求不定积分∫[1/(u²-2u)]du

∫[1/(u²-2u)]du=∫[1/u(u-2)]du=(1/2)∫[1/(u-2)-1/u]du=(1/2)[∫d(u-2)/(u-2)-∫du/u]

=(1/2)[ln(u-2)-lnu]+C=ln√[(u-2)/u)]+C

求不定积分∫[(1/u²)-2u]du

∫[(1/u²)-2u]du=-(1/u)-2u²+C

注：因为原题书写不清,搞不清楚主分数线管到哪里,故做了两个.