integral of 2^x*2^{2^x}

解答

\int 2^{x} \cdot 2^{2^{x}} d x

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

求解步骤

\int 2^{x} \cdot 2^{2^{x}} d x

使用换元积分法

= \int \frac{2 ^{u}}{ln ( 2 )} d u

提出常数:

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

= \frac{1}{ln ( 2 )} \cdot \int 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{ln ( 2 )} \cdot \frac{2 ^{u}}{ln ( 2 )}

u = 2^{x}

代回

= \frac{1}{ln ( 2 )} \cdot \frac{2 ^{2^{x}}}{ln ( 2 )}

化简

\frac{1}{ln ( 2 )} \cdot \frac{2 ^{2^{x}}}{ln ( 2 )} : \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2^{x}}

= \frac{1}{ln ^{2} ( 2 )} \cdot 2^{2^{x}}

解答补常数

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

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