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受欢迎的微积分 >

integral of (6e^{2x})/(e^{2x)+12e^x+27}

解答

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

解答

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

求解步骤

\int \frac{6 e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 6 \cdot \int \frac{e ^{2 x}}{e ^{2 x} + 12 e ^{x} + 27} d x

使用换元积分法

= 6 \cdot \int \frac{u}{u ^{2} + 12 u + 27} d u

对

u^{2} + 12 u + 27

配方:

(u + 6)^{2} - 9

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

使用换元积分法

= 6 \cdot \int \frac{v - 6}{v ^{2} - 9} d v

将

\frac{v - 6}{v ^{2} - 9}

用部份分式展开:

\frac{3}{2 ( v + 3 )} - \frac{1}{2 ( v - 3 )}

= 6 \cdot \int \frac{3}{2 ( v + 3 )} - \frac{1}{2 ( v - 3 )} d v

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

= 6 (\int \frac{3}{2 ( v + 3 )} d v - \int \frac{1}{2 ( v - 3 )} d v)

\int \frac{3}{2 ( v + 3 )} d v = \frac{3}{2} ln ∣ v + 3 ∣

\int \frac{1}{2 ( v - 3 )} d v = \frac{1}{2} ln ∣ v - 3 ∣

= 6 (\frac{3}{2} ln ∣ v + 3 ∣ - \frac{1}{2} ln ∣ v - 3 ∣)

代回

= 6 (\frac{3}{2} ln ∣ e^{x} + 6 + 3 ∣ - \frac{1}{2} ln ∣ e^{x} + 6 - 3 ∣)

化简

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

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

解答补常数

= 6 (\frac{3}{2} ln ∣ e^{x} + 9 ∣ - \frac{1}{2} ln ∣ e^{x} + 3 ∣) + C

作图

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