integral of (2e^x+4^x)

解答

\int (2 e^{x} + 4^{x}) d x

2 e^{x} + \frac{2 ^{2 x - 1}}{ln ( 2 )} + C

求解步骤

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

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

\int 2 e^{x} d x = 2 e^{x}

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

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

解答补常数

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

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