求不定积分 ∫[x^3+1]dx/[x(x-1)^3]
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解：
(x³+1)/[x(x-1)³]=2/(x-1)³-1/x+2/(x-1)+1/(x-1)²
∫(x³+1)/[x(x-1)³] dx
=∫[2/(x-1)³-1/x+2/(x-1)+1/(x-1)²]dx
=2∫1/(x-1)³ d(x-1)-∫1/x dx+∫2/(x-1) d(x-1)+∫1/(x-1)² d(x-1)
=-1/(x-1)²-ln|x|+2ln|x-1|-1/(x-1)+C
(x³+1)/[x(x-1)³]=2/(x-1)³-1/x+2/(x-1)+1/(x-1)²
∫(x³+1)/[x(x-1)³] dx
=∫[2/(x-1)³-1/x+2/(x-1)+1/(x-1)²]dx
=2∫1/(x-1)³ d(x-1)-∫1/x dx+∫2/(x-1) d(x-1)+∫1/(x-1)² d(x-1)
=-1/(x-1)²-ln|x|+2ln|x-1|-1/(x-1)+C
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