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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

以下の部分分数を得る:

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

総和規則を適用する:

\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 ∣

代用を戻す

u = x + 3

= - \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 ∣

簡素化

- \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 ∣

定数を解答に追加する

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

グラフ

人気の例

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