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受欢迎的微积分 >

integral of sqrt(e^{10x)-9}

解答

\int e^{10 x} - 9 d x

解答

- \frac{3}{5} arctan (\frac{1}{3} e^{10 x} - 9) + \frac{1}{5} e^{10 x} - 9 + C

求解步骤

\int e^{10 x} - 9 d x

使用换元积分法

= \int \frac{u}{10 ( u + 9 )} d u

提出常数:

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

= \frac{1}{10} \cdot \int \frac{u}{u + 9} d u

使用换元积分法

= \frac{1}{10} \cdot \int \frac{2 v ^{2}}{v ^{2} + 9} d v

提出常数:

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

= \frac{1}{10} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} + 9} d v

\frac{v ^{2}}{v ^{2} + 9} = - \frac{9}{v ^{2} + 9} + 1

= \frac{1}{10} \cdot 2 \cdot \int - \frac{9}{v ^{2} + 9} + 1 d v

使用换元积分法

= \frac{1}{10} \cdot 2 \cdot \int 3 (- \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}{10} \cdot 2 \cdot 3 \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}{10} \cdot 2 \cdot 3 (- \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}{10} \cdot 2 \cdot 3 (- arctan (w) + w)

代回

= \frac{1}{10} \cdot 2 \cdot 3 (- arctan (\frac{e ^{10 x} - 9}{3}) + \frac{e ^{10 x} - 9}{3})

化简

\frac{1}{10} \cdot 2 \cdot 3 (- arctan (\frac{e ^{10 x} - 9}{3}) + \frac{e ^{10 x} - 9}{3}) : - \frac{3}{5} arctan (\frac{1}{3} e^{10 x} - 9) + \frac{1}{5} e^{10 x} - 9

= - \frac{3}{5} arctan (\frac{1}{3} e^{10 x} - 9) + \frac{1}{5} e^{10 x} - 9

解答补常数

= - \frac{3}{5} arctan (\frac{1}{3} e^{10 x} - 9) + \frac{1}{5} e^{10 x} - 9 + C

作图

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