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人気のある微分積分 >

integral of 5/((x-2)(x^2-1))

解

\int \frac{5}{( x - 2 ) ( x ^{2} - 1 )} d x

解

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

解答ステップ

\int \frac{5}{( x - 2 ) ( x ^{2} - 1 )} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 5 \cdot \int \frac{1}{( x - 2 ) ( x ^{2} - 1 )} d x

u置換積分法を適用する

= 5 \cdot \int \frac{1}{u ( u ^{2} + 4 u + 3 )} d u

以下の部分分数を得る:

\frac{1}{u ( u ^{2} + 4 u + 3 )} : \frac{1}{3 u} - \frac{1}{2 ( u + 1 )} + \frac{1}{6 ( u + 3 )}

= 5 \cdot \int \frac{1}{3 u} - \frac{1}{2 ( u + 1 )} + \frac{1}{6 ( u + 3 )} 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

= 5 (\int \frac{1}{3 u} d u - \int \frac{1}{2 ( u + 1 )} d u + \int \frac{1}{6 ( u + 3 )} d u)

\int \frac{1}{3 u} d u = \frac{1}{3} ln ∣ u ∣

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

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

= 5 (\frac{1}{3} ln ∣ u ∣ - \frac{1}{2} ln ∣ u + 1 ∣ + \frac{1}{6} ln ∣ u + 3 ∣)

代用を戻す

u = x - 2

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 2 + 1 ∣ + \frac{1}{6} ln ∣ x - 2 + 3 ∣)

簡素化

5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 2 + 1 ∣ + \frac{1}{6} ln ∣ x - 2 + 3 ∣) : 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣)

定数を解答に追加する

= 5 (\frac{1}{3} ln ∣ x - 2 ∣ - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{6} ln ∣ x + 1 ∣) + C

グラフ

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