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(x-根号x 1)的不定积分是多少
1个回答
2015-11-13
展开全部
令√x=u,得:x=u^2,∴dx=2udu.∴原式=∫[u^2/(1-u)](2u)du=-2∫u^3/(u-1)d(u-1).再令u-1=v,得:u=v+1,∴u^3=(v+1)^3=v^3+3v^2+3v+1.∴原式=-2∫[(v^3+3v^2+3v+1)/v]dv=-2∫v^2dv-6∫vdv-6∫dv-2∫(1/v)dv =-(2/3)v^3-3v^2-6v-2ln|v|+C =-(2/3)(u-1)^3-3(u-1)^2-6(u-1)-2ln|u-1|+C =-(2/3)(u^3-3u^2+3u-1)-3(u^2-2u+1)-6u+6-2ln|u-1|+C =-(2/3)u^3+2u^2-2u+2/3-3u^2+6u-3-6u-2ln|u-1|+C =-(2/3)u^3-u^2-2u-2ln|u-1|+C =-(2/3)x√x-x-2√x-2ln|√x-1|+C
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