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人気のある微分積分 >

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

解

\int \frac{2 x}{4 - x ^{4}} d x

解

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

解答ステップ

\int \frac{2 x}{4 - x ^{4}} 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}{4 - x ^{4}} d x

u置換積分法を適用する

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

定数を除く:

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

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

置換積分法を適用する

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

定数を除く:

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

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \int \frac{1}{- v ^{2} + 1} d v

共通積分を使用する:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

代用を戻す

= 2 \cdot \frac{1}{2} \cdot \frac{1}{2} \frac{ln \frac{x ^{2}}{2} + 1}{2} - \frac{ln \frac{x ^{2}}{2} - 1}{2}

簡素化

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

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

定数を解答に追加する

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

グラフ

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