如图，不定积分，如何用待定系数法拆开成两项？
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1/[u(u-2)] = A/u + B/(u-2) = [(A+B)u-2A]/[u(u-2)]
比较两边u的同次幂系数，得-2A=1,A+B=0,解得A=-1/2,B=1/2
得 1/[u(u-2)] = -(1/2)/u + (1/2)/(u-2).
像本题较简单， 也可凑:
1/[u(u-2)] = (1/2){2-u+u/[u(u-2)]} = (1/2)[-1/u + 1/(u-2)]
比较两边u的同次幂系数，得-2A=1,A+B=0,解得A=-1/2,B=1/2
得 1/[u(u-2)] = -(1/2)/u + (1/2)/(u-2).
像本题较简单， 也可凑:
1/[u(u-2)] = (1/2){2-u+u/[u(u-2)]} = (1/2)[-1/u + 1/(u-2)]
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