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人気のある微分積分 >

integral of 2/(2+e^x)

解

\int \frac{2}{2 + e ^{x}} d x

解

x - ln ∣ e^{x} + 2 ∣ + C

解答ステップ

\int \frac{2}{2 + e ^{x}} 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{1}{2 + e ^{x}} d x

u置換積分法を適用する

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

以下の部分分数を得る:

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

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

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

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

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

代用を戻す

u = e^{x}

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

簡素化

2 (\frac{1}{2} ln ∣ e^{x} ∣ - \frac{1}{2} ln ∣ e^{x} + 2 ∣) : x - ln ∣ e^{x} + 2 ∣

= x - ln ∣ e^{x} + 2 ∣

定数を解答に追加する

= x - ln ∣ e^{x} + 2 ∣ + C

グラフ

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