∫\(x^{4}\)dx/(\(x^{3}\)-3x+2)dx
= ∫(x(\(x^{3}\) - 3x+2) + 3\(x^{2}\) - 2x)dx/(\(x^{3}\) - 3x+2)
= ∫(x + (3\(x^{2}\) - 2x)/(\(x^{3}\) - 3x+2))dx
= ∫xdx + ∫((3\(x^{2}\) - 2x)/(\(x^{3}\) - 3x+2))dx
= \(x^{2}\)/2 + ∫((3\(x^{2}\) - 2x)/((x + 2)(x - 1\()^{2}\))) dx

ここで
(3\(x^{2}\) - 2x)/((x + 2)(x - 1\()^{2}\)) = a/(x + 2) + b/(x - 1) + c/(x - 1\()^{2}\)
と置いて分母を払うと
a(x - 1\()^{2}\) + b(x + 2)(x - 1) + c(x + 2) = 3\(x^{2}\) - 2x
この式で
x = -2 ⇒ 9a = 16 ∴a = \(\frac{16}{9}\).
x = 1 ⇒ 3c = 1 ∴c = \(\frac{1}{3}\).
従って
16(x - 1\()^{2}\) + 9b(x + 2)(x - 1) + 3(x + 2) = 9(3\(x^{2}\) - 2x)
x = 0 を代入すると
16 - 18b + 6 = 0
∴b = \(\frac{11}{9}\).

以上より
与式 = \(x^{2}\)/2 + (\(\frac{16}{9}\))∫dx/(x + 2)
　　　　　　 + (\(\frac{11}{9}\))∫dx/(x - 1) + (\(\frac{1}{3}\))∫dx/(x -1\()^{2}\)
　　 = \(x^{2}\)/2 + (\(\frac{16}{9}\)) log |x + 2| + (\(\frac{11}{9}\)) log |x - 1|
　　　　　　 + 1/(3(x - 1)) + C,
C は積分定数。