通過待定系數法，對數、反正切函數不定積分公式，綜合計算不定積分∫(x 1)dx/[(x-1)(x^2 x 3)]。
解題步驟：設(x 1)/[(x-1)(x^2 x 3)]
=(mx n)/(x^2 x 3)-m/(x-1)
則：
-2m n=1,
-3m-n=1。
解得：m=-2/5，n=1/5。
此時不定積分為：
∫(x 1)dx/[(x-1)(x^2 x 3)]
=∫2/5dx/(x-1)-∫(2/5x-1/5)dx/(x^2 x 3)
=2/5ln|x-1|-∫[1/5(2x 1)-2/5]dx/(x^2 x 3)
本步驟用到對數函數不定積分公式∫dx/x=ln|x| C.
=2/5ln|x-1|-1/5∫d(x^2 x 3)/(x^2 x 3) 2/5∫dx/(x^2 x 3)
=2/5ln|x-1|-1/5ln(x^2 x 3) 2/5∫dx/[(x 1/2)^2 11/4]
=2/5ln|x-1|-1/5ln(x^2 x 3) [4/(5√11)]∫d(x/√11/2)/[(x 1/2)/(√11/2)]^2 1
=2/5ln|x-1|-1/5ln(x^2 x 3) [4/(5√11)]arctan[(x 1/2)/(√11/2)] C.
本步驟用到反正切函數不定積分公式∫dx/(x^2 1)=arctanx C.
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