作业帮∫(2x+2)/[(x-1)*(x^2+1)^2]dx
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∫(2x+2)/[(x-1)*(x^2+1)^2]dx
∫(2x+2)/[(x-1)*(x^2+1)^2]dx
∫(2x+2)/[(x-1)*(x^2+1)^2]dx
令(2x+2)/[(x-1)(x²+1)²]=A/(x-1) + (Bx+C)/(x²+1) + (Dx+E)/(x²+1)²
右边通分合并后,与左边比较系数得：
A=1,B=-1,C=-1,D=-2,E=0,
因此：(2x+2)/[(x-1)(x²+1)²]=1/(x-1) - (x+1)/(x²+1) - 2x/(x²+1)²
原积分=∫ 1/(x-1) dx - ∫ x/(x²+1) dx - ∫ 1/(x²+1) dx - 2∫ x/(x²+1)² dx
=ln|x-1| - (1/2)ln(x²+1) - arctanx - ∫ 1/(x²+1)² d(x²)
=ln|x-1| - (1/2)ln(x²+1) - arctanx + 1/(x²+1) + C
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