integral of (2x^2+4)/(x^3-8)

解答

\int \frac{2 x ^{2} + 4}{x ^{3} - 8} d x

\frac{2}{3} ln x^{3} - 8 + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

求解步骤

\int \frac{2 x ^{2} + 4}{x ^{3} - 8} d x

乘开

\frac{2 x ^{2} + 4}{x ^{3} - 8} : \frac{2 x ^{2}}{x ^{3} - 8} + \frac{4}{x ^{3} - 8}

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{2 x ^{2}}{x ^{3} - 8} d x + \int \frac{4}{x ^{3} - 8} d x

\int \frac{2 x ^{2}}{x ^{3} - 8} d x = \frac{2}{3} ln x^{3} - 8

\int \frac{4}{x ^{3} - 8} d x = \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

= \frac{2}{3} ln x^{3} - 8 + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

解答补常数

= \frac{2}{3} ln x^{3} - 8 + \frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{6} ln x^{2} + 2 x + 4 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C