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人気のある微分積分 >

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

解

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

解

2 (- \frac{1}{x - 2} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )})) + C

解答ステップ

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

定数を除く:

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

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

u置換積分法を適用する

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

拡張

\frac{u - 2}{u ( u - 4 ) ^{2}} : \frac{u}{u ( u - 4 ) ^{2}} - \frac{2}{u ( u - 4 ) ^{2}}

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

= 2 (\int \frac{u}{u ( u - 4 ) ^{2}} d u - \int \frac{2}{u ( u - 4 ) ^{2}} d u)

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

\int \frac{2}{u ( u - 4 ) ^{2}} d u = 2 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{4 ( u - 4 )})

= 2 (- \frac{1}{u - 4} - 2 (\frac{1}{16} ln ∣ u ∣ - \frac{1}{16} ln ∣ u - 4 ∣ - \frac{1}{4 ( u - 4 )}))

代用を戻す

u = x + 2

= 2 (- \frac{1}{x + 2 - 4} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x + 2 - 4 ∣ - \frac{1}{4 ( x + 2 - 4 )}))

簡素化

2 (- \frac{1}{x + 2 - 4} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x + 2 - 4 ∣ - \frac{1}{4 ( x + 2 - 4 )})) : 2 (- \frac{1}{x - 2} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )}))

= 2 (- \frac{1}{x - 2} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )}))

定数を解答に追加する

= 2 (- \frac{1}{x - 2} - 2 (\frac{1}{16} ln ∣ x + 2 ∣ - \frac{1}{16} ln ∣ x - 2 ∣ - \frac{1}{4 ( x - 2 )})) + C

グラフ

人気の例

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

integral of (4x+3)(ln(2x-3))

\int (4 x + 3) (ln (2 x - 3)) d x

integral of (2sin(1/x))/(x^2)

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

integral of (ln(ln(k)))/(2k)

\int \frac{ln ( ln ( k ) )}{2 k} d k

integral of (10z+2)/((z+1)^2(z-3)^2)

\int \frac{10 z + 2}{( z + 1 ) ^{2} ( z - 3 ) ^{2}} d z