求积分1/(x^2+1)(x^2+x) dx 假设1/(x^2+1)(x^2+x)＝(ax+b)/(x^2+1)+c/x

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求积分1/(x^2+1)(x^2+x) dx 假设1/(x^2+1)(x^2+x)＝(ax+b)/(x^2+1)+c/x+d/(x+1) 怎么求a,b,c

假设1/(x²+1)(x²+x)＝(ax+b)/(x²+1)+c/x+d/(x+1)
等式右边=((ax+b)(x²+x)+c(x²+1)(x+1)+dx(x²+1))/(x²+1)(x²+x)
=((a+c+d)x³+(a+b+c)x²+(b+c+d)x+c)/(x²+1)(x²+x)
对应项系数相等
∴a+c+d=0 a+b+c=0 b+c+d=0 c=1
解得a=-1/2 b=-1/2 c=1 d=-1/2

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