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人気のある微分積分 >

integral of 8/((x-2)(x^2+4))

解

\int \frac{8}{( x - 2 ) ( x ^{2} + 4 )} d x

解

ln \frac{x}{2} - 1 - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2}) + C

解答ステップ

\int \frac{8}{( x - 2 ) ( x ^{2} + 4 )} d x

定数を除く:

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

= 8 \cdot \int \frac{1}{( x - 2 ) ( x ^{2} + 4 )} d x

置換積分法を適用する

= 8 \cdot \int \frac{1}{4 ( u - 1 ) ( u ^{2} + 1 )} d u

定数を除く:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{( u - 1 ) ( u ^{2} + 1 )} d u

以下の部分分数を得る:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{2 ( u - 1 )} + \frac{- u - 1}{2 ( u ^{2} + 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

= 8 \cdot \frac{1}{4} (\int \frac{1}{2 ( u - 1 )} d u + \int \frac{- u - 1}{2 ( u ^{2} + 1 )} d u)

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

\int \frac{- u - 1}{2 ( u ^{2} + 1 )} d u = \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u))

= 8 \cdot \frac{1}{4} (\frac{1}{2} ln ∣ u - 1 ∣ + \frac{1}{2} (- \frac{1}{2} ln u^{2} + 1 - arctan (u)))

代用を戻す

u = \frac{x}{2}

= 8 \cdot \frac{1}{4} (\frac{1}{2} ln \frac{x}{2} - 1 + \frac{1}{2} (- \frac{1}{2} ln (\frac{x}{2})^{2} + 1 - arctan (\frac{x}{2})))

簡素化

8 \cdot \frac{1}{4} (\frac{1}{2} ln \frac{x}{2} - 1 + \frac{1}{2} (- \frac{1}{2} ln (\frac{x}{2})^{2} + 1 - arctan (\frac{x}{2}))) : ln \frac{x}{2} - 1 - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2})

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

定数を解答に追加する

= ln \frac{x}{2} - 1 - \frac{1}{2} ln \frac{x ^{2}}{4} + 1 - arctan (\frac{x}{2}) + C

グラフ

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