求(x^2+1)/(x^2-2x+1)^2 不定积分

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/15 05:23:18

原式 =∫(x^2+1)dx/(x-1)^4
=∫[1/(x-1)^2+2/(x-1)^3+2/(x-1)^4]d(x-1)
= -1/(x-1)-1/(x-1)^2-(2/3)/(x-1)^3 + C
再问：抱歉，分母常数为2
再答： I = ∫(x^2+1)dx/(x^2-2x+2)^2 = ∫(x^2+1)dx/[(x-1)^2+1]^2,
令 x-1= tant，则 x=1+tant， dx=(sect)^2dt,
I = ∫[(tant)^2+2tant+2]dt/(sect)^2
= ∫[(sint)^2+sin2t+2(cost)^2]dt
= ∫[1+sin2t+(cost)^2]dt = ∫[3/2+sin2t+(1/2)cos2t]dt
= 3t/2-(1/2)cos2t+(1/4)sin2t+C
= 3t/2-(1/2)[(cost)^2-(sint)^]+(1/2)sintcost+C
= (3/2)arctan(x-1)+(1/2)(x^2-x-1)/(x^2-2x+2)+C

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