有理函数的积分3

不定积分~1∫dx/(x^2+1

解：1。谈不上简单，只能说是经验 1/[(x²+1)(x²+x+1)] =(1/x)[1/(x²+1)-1/(x²+x+1)] =1/[x(x²+1)]-1/[x(x²+x+1)] =(1/x)-x/(x²+1)-[(1/x)-(x+1)/(x²+x+1)] =(x+1)/(x²+x+1)-x/(x²+1) 故原积分=∫(x+1)dx/(x²+x+1)-∫xdx/(x²+1) =∫(x+1/2)dx/(x²+x+1)+(1/2)∫dx/(x²+x+1)...全部

  解：1。谈不上简单，只能说是经验 1/[(x²+1)(x²+x+1)] =(1/x)[1/(x²+1)-1/(x²+x+1)] =1/[x(x²+1)]-1/[x(x²+x+1)] =(1/x)-x/(x²+1)-[(1/x)-(x+1)/(x²+x+1)] =(x+1)/(x²+x+1)-x/(x²+1) 故原积分=∫(x+1)dx/(x²+x+1)-∫xdx/(x²+1) =∫(x+1/2)dx/(x²+x+1)+(1/2)∫dx/(x²+x+1) -(1/2)∫dx²/(x²+1) =(1/2)∫d(x²+x+1)/(x²+x+1)+(1/2)∫dx/[(x+1/2)²+(3/4)] -(1/2)∫d(x²+1)/(x²+1) =(1/2)ln(x²+x+1)+(1/2)arctan[(x+1/2)/(√3/2)] -(1/2)ln(x²+1)+C =(1/2)ln(x²+x+1)+(1/2)arctan(2x+1/√3)-(1/2)ln(x²+1)+C 2。
  要熟悉∫(1+x²)dx/(x^4+1)和∫(1-x²)dx/(x^4+1)的积分方法才能想到用下面的求法 ∫dx/(x^4+1) =(1/2)∫[(1+x²)+(1-x²)]dx/(x^4+1) =(1/2)∫(1+x²)dx/(x^4+1)+(1/2)∫(1-x²)dx/(x^4+1) =(1/2)∫[1+(1/x²)]dx/[x²+(1/x²)](分子分母同除以x²) -(1/2)∫[1-(1/x²)]dx/[x²+(1/x²)] =(1/2)∫d[x-(1/x)]/{[x-(1/x)]²+2} -(1/2)∫d[x+(1/x)]/{[x+(1/x)]²-2} =(1/2√2)arctan{[x-(1/x)]/√2} -(1/4√2)ln|[x+(1/x)-√2]/[x+(1/x)+√2]|+C =(√2/4)arctan[(x²-1)/x√2} -(√2/8)ln|(x²-x√2+1)/(x²+x√2+1)|+C。
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