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人気のある微分積分 >

integral of 1/(e^x-9e^{-x)}

解

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

解

- \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

解答ステップ

\int \frac{1}{e ^{x} - 9 e ^{- x}} d x

因数

e^{x} - 9 e^{- x} : e^{- x} (e^{x} + 3) (e^{x} - 3)

= \int \frac{1}{e ^{- x} ( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

指数の規則を適用する:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{e ^{- x} ( e ^{x} + 3 ) ( e ^{x} - 3 )} = e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )}

= \int e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

簡素化

e^{x} \frac{1}{( e ^{x} + 3 ) ( e ^{x} - 3 )} : \frac{e ^{x}}{( e ^{x} + 3 ) ( e ^{x} - 3 )}

= \int \frac{e ^{x}}{( e ^{x} + 3 ) ( e ^{x} - 3 )} d x

u置換積分法を適用する

= \int \frac{1}{( u + 3 ) ( u - 3 )} d u

以下の部分分数を得る:

\frac{1}{( u + 3 ) ( u - 3 )} : - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )}

= \int - \frac{1}{6 ( u + 3 )} + \frac{1}{6 ( u - 3 )} 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}{6 ( u + 3 )} d u + \int \frac{1}{6 ( u - 3 )} d u

\int \frac{1}{6 ( u + 3 )} d u = \frac{1}{6} ln ∣ u + 3 ∣

\int \frac{1}{6 ( u - 3 )} d u = \frac{1}{6} ln ∣ u - 3 ∣

= - \frac{1}{6} ln ∣ u + 3 ∣ + \frac{1}{6} ln ∣ u - 3 ∣

代用を戻す

u = e^{x}

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣

定数を解答に追加する

= - \frac{1}{6} ln ∣ e^{x} + 3 ∣ + \frac{1}{6} ln ∣ e^{x} - 3 ∣ + C

グラフ

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