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人気のある微分積分 >

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

解

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

解

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

解答ステップ

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

平方完成する

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

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

u置換積分法を適用する

= 2 \cdot \int \frac{2 u + 1}{8 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}{8} \cdot \int \frac{2 u + 1}{u ^{2}} d u

拡張

\frac{2 u + 1}{u ^{2}} : \frac{2}{u} + \frac{1}{u ^{2}}

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

\int \frac{2}{u} d u = 2 ln ∣ u ∣

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

= 2 \cdot \frac{1}{8} (2 ln ∣ u ∣ - \frac{1}{u})

代用を戻す

u = x - \frac{1}{2}

= 2 \cdot \frac{1}{8} (2 ln x - \frac{1}{2} - \frac{1}{x - \frac{1}{2}})

簡素化

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

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

定数を解答に追加する

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

グラフ

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