integral of x/(x^3-x^2+x+3)

解答

\int \frac{x}{x ^{3} - x ^{2} + x + 3} d x

- \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 3 + 42 arctan (\frac{1}{2} (x - 1))) + C

求解步骤

\int \frac{x}{x ^{3} - x ^{2} + x + 3} d x

将

\frac{x}{x ^{3} - x ^{2} + x + 3}

用部份分式展开:

- \frac{1}{6 ( x + 1 )} + \frac{x + 3}{6 ( x ^{2} - 2 x + 3 )}

= \int - \frac{1}{6 ( x + 1 )} + \frac{x + 3}{6 ( x ^{2} - 2 x + 3 )} d x

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{6 ( x + 1 )} d x + \int \frac{x + 3}{6 ( x ^{2} - 2 x + 3 )} d x

\int \frac{1}{6 ( x + 1 )} d x = \frac{1}{6} ln ∣ x + 1 ∣

\int \frac{x + 3}{6 ( x ^{2} - 2 x + 3 )} d x = \frac{1}{12} (ln x^{2} - 2 x + 3 + 42 arctan (\frac{1}{2} (x - 1)))

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 3 + 42 arctan (\frac{1}{2} (x - 1)))

解答补常数

= - \frac{1}{6} ln ∣ x + 1 ∣ + \frac{1}{12} (ln x^{2} - 2 x + 3 + 42 arctan (\frac{1}{2} (x - 1))) + C

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