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人気のある微分積分 >

integral of (4x)/((x-1)(x^2-1))

解

\int \frac{4 x}{( x - 1 ) ( x ^{2} - 1 )} d x

解

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

解答ステップ

\int \frac{4 x}{( x - 1 ) ( x ^{2} - 1 )} d x

定数を除く:

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

= 4 \cdot \int \frac{x}{( x - 1 ) ( x ^{2} - 1 )} d x

以下の部分分数を得る:

\frac{x}{( x - 1 ) ( x ^{2} - 1 )} : \frac{1}{4 ( x - 1 )} + \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{4 ( x + 1 )}

= 4 \cdot \int \frac{1}{4 ( x - 1 )} + \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{4 ( x + 1 )} d x

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

= 4 (\int \frac{1}{4 ( x - 1 )} d x + \int \frac{1}{2 ( x - 1 ) ^{2}} d x - \int \frac{1}{4 ( x + 1 )} d x)

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

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

\int \frac{1}{4 ( x + 1 )} d x = \frac{1}{4} ln ∣ x + 1 ∣

= 4 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣)

簡素化

4 (\frac{1}{4} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} - \frac{1}{4} ln ∣ x + 1 ∣) : ln ∣ x - 1 ∣ - \frac{2}{x - 1} - ln ∣ x + 1 ∣

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

定数を解答に追加する

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

グラフ

人気の例

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