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人気のある微分積分 >

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

解

\int \frac{e ^{x}}{( e ^{x} - 1 ) ( e ^{x} + 4 )} d x

解

\frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣ + C

解答ステップ

\int \frac{e ^{x}}{( e ^{x} - 1 ) ( e ^{x} + 4 )} d x

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{( u - 1 ) ( u + 4 )} : \frac{1}{5 ( u - 1 )} - \frac{1}{5 ( u + 4 )}

= \int \frac{1}{5 ( u - 1 )} - \frac{1}{5 ( u + 4 )} 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

= \int \frac{1}{5 ( u - 1 )} d u - \int \frac{1}{5 ( u + 4 )} d u

\int \frac{1}{5 ( u - 1 )} d u = \frac{1}{5} ln ∣ u - 1 ∣

\int \frac{1}{5 ( u + 4 )} d u = \frac{1}{5} ln ∣ u + 4 ∣

= \frac{1}{5} ln ∣ u - 1 ∣ - \frac{1}{5} ln ∣ u + 4 ∣

代用を戻す

u = e^{x}

= \frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣

定数を解答に追加する

= \frac{1}{5} ln ∣ e^{x} - 1 ∣ - \frac{1}{5} ln ∣ e^{x} + 4 ∣ + C

グラフ

人気の例

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