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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

以下の部分分数を得る:

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

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

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

\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 )}

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

簡素化

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

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

定数を解答に追加する

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

グラフ

人気の例

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\int 1 + sin (x) cos (x) d x

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\int x + \frac{2}{sin ( x )} d x

integral of x\sqrt[4]{3x-2}

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integral of (x+1)^3(x-2)

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integral of 1/x sqrt(25-x^2)

\int \frac{1}{x} 25 - x^{2} d x