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人気のある微分積分 >

integral of (e^x)/((e^x-8)(e^{2x)+1)}

解

\int \frac{e ^{x}}{( e ^{x} - 8 ) ( e ^{2 x} + 1 )} d x

解

\frac{1}{65} ln ∣ e^{x} - 8 ∣ + \frac{1}{65} (- \frac{1}{2} ln e^{2 x} + 1 - 8 arctan (e^{x})) + C

解答ステップ

\int \frac{e ^{x}}{( e ^{x} - 8 ) ( e ^{2 x} + 1 )} d x

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{( u - 8 ) ( u ^{2} + 1 )} : \frac{1}{65 ( u - 8 )} + \frac{- u - 8}{65 ( u ^{2} + 1 )}

= \int \frac{1}{65 ( u - 8 )} + \frac{- u - 8}{65 ( u ^{2} + 1 )} 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}{65 ( u - 8 )} d u + \int \frac{- u - 8}{65 ( u ^{2} + 1 )} d u

\int \frac{1}{65 ( u - 8 )} d u = \frac{1}{65} ln ∣ u - 8 ∣

\int \frac{- u - 8}{65 ( u ^{2} + 1 )} d u = \frac{1}{65} (- \frac{1}{2} ln u^{2} + 1 - 8 arctan (u))

= \frac{1}{65} ln ∣ u - 8 ∣ + \frac{1}{65} (- \frac{1}{2} ln u^{2} + 1 - 8 arctan (u))

代用を戻す

u = e^{x}

= \frac{1}{65} ln ∣ e^{x} - 8 ∣ + \frac{1}{65} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 8 arctan (e^{x}))

指数の規則を適用する:

(a^{b})^{c} = a^{b c}

= \frac{1}{65} ln ∣ e^{x} - 8 ∣ + \frac{1}{65} (- \frac{1}{2} ln e^{2 x} + 1 - 8 arctan (e^{x}))

定数を解答に追加する

= \frac{1}{65} ln ∣ e^{x} - 8 ∣ + \frac{1}{65} (- \frac{1}{2} ln e^{2 x} + 1 - 8 arctan (e^{x})) + C

グラフ

人気の例

limit as x approaches infinity of 3+5/x

x \to \infty lim (3 + \frac{5}{x})

(\partial)/(\partial x)(d/t)

\frac{\partial}{\partial x} (\frac{d}{t})

integral from 0 to 5 of |5x-10|

\int_{0}^{5} ∣ 5 x - 10 ∣ d x

y^'=3x(y-1)^{1/3}

y^{^{'}} = 3 x (y - 1)^{\frac{1}{3}}

面積 10x-5x^2,x=\sqrt[3]{4},x=2

a re a 10 x - 5 x^{2}, x = 34, x = 2