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人気のある微分積分 >

integral of (e^x)/((e^x-9)(e^{2x)+1)}

解

\int \frac{e ^{x}}{( e ^{x} - 9 ) ( e ^{2 x} + 1 )} d x

解

\frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x})) + C

解答ステップ

\int \frac{e ^{x}}{( e ^{x} - 9 ) ( e ^{2 x} + 1 )} d x

u置換積分法を適用する

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

以下の部分分数を得る:

\frac{1}{( u - 9 ) ( u ^{2} + 1 )} : \frac{1}{82 ( u - 9 )} + \frac{- u - 9}{82 ( u ^{2} + 1 )}

= \int \frac{1}{82 ( u - 9 )} + \frac{- u - 9}{82 ( 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}{82 ( u - 9 )} d u + \int \frac{- u - 9}{82 ( u ^{2} + 1 )} d u

\int \frac{1}{82 ( u - 9 )} d u = \frac{1}{82} ln ∣ u - 9 ∣

\int \frac{- u - 9}{82 ( u ^{2} + 1 )} d u = \frac{1}{82} (- \frac{1}{2} ln u^{2} + 1 - 9 arctan (u))

= \frac{1}{82} ln ∣ u - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln u^{2} + 1 - 9 arctan (u))

代用を戻す

u = e^{x}

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 9 arctan (e^{x}))

指数の規則を適用する:

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

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x}))

定数を解答に追加する

= \frac{1}{82} ln ∣ e^{x} - 9 ∣ + \frac{1}{82} (- \frac{1}{2} ln e^{2 x} + 1 - 9 arctan (e^{x})) + C

グラフ

人気の例

derivative of 1/6

\frac{d}{d x} (\frac{1}{6})

derivative (1+4x^2)(x-x^2)

d er i v a t i v e (1 + 4 x^{2}) (x - x^{2})

derivative of Acos(x+Bsin(x))

\frac{d}{d x} (A cos (x) + B sin (x))

integral from-1 to 6 of-x^2+5x+6

\int_{- 1}^{6} - x^{2} + 5 x + 6 d x

面積 y= 1/2 x,y= 6/x

a re a y = \frac{1}{2} x, y = \frac{6}{x}