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受欢迎的微积分 >

integral of e^{2x}sqrt(1+4e^x)

解答

\int e^{2 x} 1 + 4 e^{x} d x

解答

\frac{1}{60} (6 e^{x} - 1) (4 e^{x} + 1)^{\frac{3}{2}} + C

求解步骤

\int e^{2 x} 1 + 4 e^{x} d x

使用换元积分法

= \int \frac{u ( u - 1 )}{16} d u

提出常数:

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

= \frac{1}{16} \cdot \int u (u - 1) d u

乘开

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{16} \cdot \int u^{\frac{3}{2}} - u 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

= \frac{1}{16} (\int u^{\frac{3}{2}} d u - \int u d u)

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{16} (\frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}})

u = 1 + 4 e^{x}

代回

= \frac{1}{16} (\frac{2}{5} (1 + 4 e^{x})^{\frac{5}{2}} - \frac{2}{3} (1 + 4 e^{x})^{\frac{3}{2}})

化简

\frac{1}{16} (\frac{2}{5} (1 + 4 e^{x})^{\frac{5}{2}} - \frac{2}{3} (1 + 4 e^{x})^{\frac{3}{2}}) : \frac{1}{60} (6 e^{x} - 1) (4 e^{x} + 1)^{\frac{3}{2}}

= \frac{1}{60} (6 e^{x} - 1) (4 e^{x} + 1)^{\frac{3}{2}}

解答补常数

= \frac{1}{60} (6 e^{x} - 1) (4 e^{x} + 1)^{\frac{3}{2}} + C

作图

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