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受欢迎的微积分 >

integral of 4^x*e^{2x}

解答

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

解答

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

求解步骤

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

使用换元积分法

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

使用换元积分法

= \frac{1}{ln ( 2 )} \cdot \int \frac{v ^{\frac{1}{l n ( 2 )}}}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} d v

提出常数:

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

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} \cdot \int v^{\frac{1}{l n ( 2 )}} d v

使用幂法则

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} \cdot \frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}

代回

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} \cdot \frac{1}{1 + ln ( 2 )} ln (2) (2^{\frac{l n ( 4 ^{x} ) + l n ( 2 )}{l n ( 2 )}})^{\frac{1 + l n ( 2 )}{l n ( 2 )}}

化简

\frac{1}{ln ( 2 )} \cdot \frac{1}{4 \cdot 2 ^{\frac{1}{l n ( 2 )}}} \cdot \frac{1}{1 + ln ( 2 )} ln (2) (2^{\frac{l n ( 4 ^{x} ) + l n ( 2 )}{l n ( 2 )}})^{\frac{1 + l n ( 2 )}{l n ( 2 )}} : \frac{1}{2 ^{\frac{1}{l n ( 2 )}} ( 1 + ln ( 2 ) )} \cdot 2^{\frac{2 x + 1 + 2 l n ( 2 ) x - l n ( 2 )}{l n ( 2 )}}

= \frac{1}{2 ^{\frac{1}{l n ( 2 )}} ( 1 + ln ( 2 ) )} \cdot 2^{\frac{2 x + 1 + 2 l n ( 2 ) x - l n ( 2 )}{l n ( 2 )}}

解答补常数

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

作图

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