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受欢迎的微积分 >

integral of 1/(x(x+1)(x+2)(x+3))

解答

\int \frac{1}{x ( x + 1 ) ( x + 2 ) ( x + 3 )} d x

解答

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

求解步骤

\int \frac{1}{x ( x + 1 ) ( x + 2 ) ( x + 3 )} d x

使用换元积分法

= \int \frac{1}{u ( u - 3 ) ( u - 2 ) ( u - 1 )} d u

将

\frac{1}{u ( u - 3 ) ( u - 2 ) ( u - 1 )}

用部份分式展开:

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

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

\int \frac{1}{6 u} d u = \frac{1}{6} ln ∣ u ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

\int \frac{1}{2 ( u - 2 )} d u = \frac{1}{2} ln ∣ u - 2 ∣

\int \frac{1}{2 ( u - 1 )} d u = \frac{1}{2} ln ∣ u - 1 ∣

= - \frac{1}{6} ln ∣ u ∣ + \frac{1}{6} ln ∣ u - 3 ∣ - \frac{1}{2} ln ∣ u - 2 ∣ + \frac{1}{2} ln ∣ u - 1 ∣

u = x + 3

代回

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

化简

- \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x + 3 - 3 ∣ - \frac{1}{2} ln ∣ x + 3 - 2 ∣ + \frac{1}{2} ln ∣ x + 3 - 1 ∣ : - \frac{1}{6} ln ∣ x + 3 ∣ + \frac{1}{6} ln ∣ x ∣ - \frac{1}{2} ln ∣ x + 1 ∣ + \frac{1}{2} ln ∣ x + 2 ∣

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

解答补常数

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

作图

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