integral of (-3x^3+2x^2-3x-6)/(x^3+x)

解答

\int \frac{- 3 x ^{3} + 2 x ^{2} - 3 x - 6}{x ^{3} + x} d x

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

求解步骤

\int \frac{- 3 x ^{3} + 2 x ^{2} - 3 x - 6}{x ^{3} + x} d x

乘开

\frac{- 3 x ^{3} + 2 x ^{2} - 3 x - 6}{x ^{3} + x} : - \frac{3 x ^{3}}{x ^{3} + x} + \frac{2 x ^{2}}{x ^{3} + x} - \frac{3 x}{x ^{3} + x} - \frac{6}{x ^{3} + 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{3 x ^{3}}{x ^{3} + x} d x + \int \frac{2 x ^{2}}{x ^{3} + x} d x - \int \frac{3 x}{x ^{3} + x} d x - \int \frac{6}{x ^{3} + x} d x

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

\int \frac{2 x ^{2}}{x ^{3} + x} d x = ln x^{2} + 1

\int \frac{3 x}{x ^{3} + x} d x = 3 arctan (x)

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

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

化简

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

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

解答补常数

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

t an g e n t \frac{1}{3} x^{3} - \frac{16}{x} - \frac{40}{3}

h \to 0 lim (7)

\int (8 x^{3} - 9 x^{2} + 4) d x

d er i v a t i v e 5 x - 1

n = 0 \sum \infty \frac{3 ^{n}}{4 ^{n}}