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人気のある微分積分 >

integral of 1/((e^x-3)^2)

解

\int \frac{1}{( e ^{x} - 3 ) ^{2}} d x

解

\frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )} + C

解答ステップ

\int \frac{1}{( e ^{x} - 3 ) ^{2}} d x

e^{x} - 3 = (e^{\frac{x}{2}} + 3) (e^{\frac{x}{2}} - 3)

= \int \frac{1}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} d x

簡素化

\frac{1}{( ( e ^{\frac{x}{2}} + 3 ) ( e ^{\frac{x}{2}} - 3 ) ) ^{2}} : \frac{1}{- 6 e ^{x} + e ^{2 x} + 9}

= \int \frac{1}{- 6 e ^{x} + e ^{2 x} + 9} d x

u置換積分法を適用する

= \int \frac{1}{u ( - 6 u + u ^{2} + 9 )} d u

以下の部分分数を得る:

\frac{1}{u ( - 6 u + u ^{2} + 9 )} : \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{2}}

= \int \frac{1}{9 u} - \frac{1}{9 ( u - 3 )} + \frac{1}{3 ( u - 3 ) ^{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

= \int \frac{1}{9 u} d u - \int \frac{1}{9 ( u - 3 )} d u + \int \frac{1}{3 ( u - 3 ) ^{2}} d u

\int \frac{1}{9 u} d u = \frac{1}{9} ln ∣ u ∣

\int \frac{1}{9 ( u - 3 )} d u = \frac{1}{9} ln ∣ u - 3 ∣

\int \frac{1}{3 ( u - 3 ) ^{2}} d u = - \frac{1}{3 ( u - 3 )}

= \frac{1}{9} ln ∣ u ∣ - \frac{1}{9} ln ∣ u - 3 ∣ - \frac{1}{3 ( u - 3 )}

代用を戻す

u = e^{x}

= \frac{1}{9} ln ∣ e^{x} ∣ - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )}

\frac{1}{9} ln ∣ e^{x} ∣ = \frac{1}{9} x

= \frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )}

定数を解答に追加する

= \frac{1}{9} x - \frac{1}{9} ln ∣ e^{x} - 3 ∣ - \frac{1}{3 ( e ^{x} - 3 )} + C

グラフ

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