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求不定积分（x+1）/(x^2+1)^2的不定积分求（x+1）/

2011-03-19 12:32:15彬***
求（x+1）/(x^2+1)^2的不定积分求不定积分（x+1）/(x^2+1)^2的不定积分求（x+1）/(x^2+1)^2的不定积分：求（x+1）/(x^2+1)^2的不定积分 ∫[(x+1)/?

最佳回答

  •   求（x+1）/(x^2+1)^2的不定积分 ∫[(x+1)/(x^2+1)^2]dx 令x=tant，则：dx=d(tant)=sec^2 tdt 原积分=∫[(tant+1)/sec^4 t]*sec^2 tdt =∫[(tant+1)/sec^2 t]dt =∫{[(sint/cost)+1]/(1/cos^2 t)}dt =∫(sintcost+cos^2 t)dt =∫sintcostdt+∫cos^2 tdt =∫sintd(sint)+(1/2)∫(cos2t+1)dt =(1/2)*(sint)^2+(1/2)[∫cos2tdt+t] =(1/2)*(sint)^2+(1/2)*[(1/2)sin2t+t]+C =(1/2)*(sint)^2+(1/2)*sintcost+(1/2)t+C =(1/2)*[(x^2+x)/(x^2+1)+arctanx]+C。
      
    2011-03-19 13:08:09
  • 1/2 ((-1 + x)/(1 + x^2) + ArcTan[x])+C(用Mathematica算的)
    2011-03-19 18:10:50
  • x=tant=sint/cost; ∫（x+1）/(x^2+1)^2dx =∫（sint+cost)/cost*(1/cost)^2/(1/cost)^4dt; =∫(sintcost+cost*cost)dt =∫(1/2)sin2tdt+∫(1/2）(1+cos2t)dt; =1/4(sin2t-cos2t)+t/2+C; =√2/4sin(2t-π/4)+t/2+C; =√2/4sin(2*(arctanx)-π/4)+arctanx/2+C;
    2011-03-19 13:02:14
    • 很赞哦！ (211)

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