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受欢迎的微积分 >

integral of (e^x)/((e^x-5)(e^{2x)+1)}

解答

\int \frac{e ^{x}}{( e ^{x} - 5 ) ( e ^{2 x} + 1 )} d x

解答

\frac{1}{26} ln ∣ e^{x} - 5 ∣ + \frac{1}{26} (- \frac{1}{2} ln e^{2 x} + 1 - 5 arctan (e^{x})) + C

求解步骤

\int \frac{e ^{x}}{( e ^{x} - 5 ) ( e ^{2 x} + 1 )} d x

使用换元积分法

= \int \frac{1}{( u - 5 ) ( u ^{2} + 1 )} d u

将

\frac{1}{( u - 5 ) ( u ^{2} + 1 )}

用部份分式展开:

\frac{1}{26 ( u - 5 )} + \frac{- u - 5}{26 ( u ^{2} + 1 )}

= \int \frac{1}{26 ( u - 5 )} + \frac{- u - 5}{26 ( u ^{2} + 1 )} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{26 ( u - 5 )} d u + \int \frac{- u - 5}{26 ( u ^{2} + 1 )} d u

\int \frac{1}{26 ( u - 5 )} d u = \frac{1}{26} ln ∣ u - 5 ∣

\int \frac{- u - 5}{26 ( u ^{2} + 1 )} d u = \frac{1}{26} (- \frac{1}{2} ln u^{2} + 1 - 5 arctan (u))

= \frac{1}{26} ln ∣ u - 5 ∣ + \frac{1}{26} (- \frac{1}{2} ln u^{2} + 1 - 5 arctan (u))

u = e^{x}

代回

= \frac{1}{26} ln ∣ e^{x} - 5 ∣ + \frac{1}{26} (- \frac{1}{2} ln (e^{x})^{2} + 1 - 5 arctan (e^{x}))

使用指数法则:

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

= \frac{1}{26} ln ∣ e^{x} - 5 ∣ + \frac{1}{26} (- \frac{1}{2} ln e^{2 x} + 1 - 5 arctan (e^{x}))

解答补常数

= \frac{1}{26} ln ∣ e^{x} - 5 ∣ + \frac{1}{26} (- \frac{1}{2} ln e^{2 x} + 1 - 5 arctan (e^{x})) + C

作图

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6y^{''}+y^'-2y=0,y(0)=5,y^'(0)=-1

6 y^{^{''}} + y^{^{'}} - 2 y = 0, y (0) = 5, y^{^{'}} (0) = - 1