integral of (x^3)/(x-1)

解答

\int \frac{x ^{3}}{x - 1} d x

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

求解步骤

\int \frac{x ^{3}}{x - 1} d x

使用换元积分法

= \int \frac{( u + 1 ) ^{3}}{u} d u

乘开

\frac{( u + 1 ) ^{3}}{u} : u^{2} + 3 u + 3 + \frac{1}{u}

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

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 u^{2} d u + \int 3 u d u + \int 3 d u + \int \frac{1}{u} d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣

u = x - 1

代回

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

解答补常数

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

(8 - 6 y + e^{- 3 x}) d x - 2 d y = 0

\frac{d}{d x} ((3 x - x^{3} + 1)^{4})

\int (40 - \frac{2}{sec ( θ )}) d θ

(5 + x^{8}) \frac{d y}{d x} = \frac{x ^{7}}{y}

x \to 2 lim (3)