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受欢迎的微积分 >

integral of sqrt(e^{6x)-4}

解答

\int e^{6 x} - 4 d x

解答

- \frac{2}{3} arctan (\frac{1}{2} e^{6 x} - 4) + \frac{1}{3} e^{6 x} - 4 + C

求解步骤

\int e^{6 x} - 4 d x

使用换元积分法

= \int \frac{u}{6 ( u + 4 )} d u

提出常数:

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

= \frac{1}{6} \cdot \int \frac{u}{u + 4} d u

使用换元积分法

= \frac{1}{6} \cdot \int \frac{2 v ^{2}}{v ^{2} + 4} d v

提出常数:

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

= \frac{1}{6} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} + 4} d v

\frac{v ^{2}}{v ^{2} + 4} = - \frac{4}{v ^{2} + 4} + 1

= \frac{1}{6} \cdot 2 \cdot \int - \frac{4}{v ^{2} + 4} + 1 d v

使用换元积分法

= \frac{1}{6} \cdot 2 \cdot \int 2 (- \frac{1}{w ^{2} + 1} + 1) d w

提出常数:

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

= \frac{1}{6} \cdot 2 \cdot 2 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

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

= \frac{1}{6} \cdot 2 \cdot 2 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

\int 1 d w = w

= \frac{1}{6} \cdot 2 \cdot 2 (- arctan (w) + w)

代回

= \frac{1}{6} \cdot 2 \cdot 2 (- arctan (\frac{e ^{6 x} - 4}{2}) + \frac{e ^{6 x} - 4}{2})

化简

\frac{1}{6} \cdot 2 \cdot 2 (- arctan (\frac{e ^{6 x} - 4}{2}) + \frac{e ^{6 x} - 4}{2}) : - \frac{2}{3} arctan (\frac{1}{2} e^{6 x} - 4) + \frac{1}{3} e^{6 x} - 4

= - \frac{2}{3} arctan (\frac{1}{2} e^{6 x} - 4) + \frac{1}{3} e^{6 x} - 4

解答补常数

= - \frac{2}{3} arctan (\frac{1}{2} e^{6 x} - 4) + \frac{1}{3} e^{6 x} - 4 + C

作图

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