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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

u置換積分法を適用する

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

拡張

\frac{u + 1}{u} : 1 + \frac{1}{u}

= 2 \cdot \frac{1}{4} \cdot \int 1 + \frac{1}{u} 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

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

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{4} (u + ln ∣ u ∣)

代用を戻す

u = 2 x - 1

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

簡素化

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

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

定数を解答に追加する

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

グラフ

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