What is the integral of 4^xe^{2x} ?

The integral of 4^xe^{2x} is 1/(2^{(1){ln(2)}}(1+ln(2))}*2^{\frac{2x+1+2ln(2)x-ln(2))/(ln(2))}+C

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integral of 4^xe^{2x}

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

\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

Solution steps

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

Apply u-substitution

: \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

= \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

Take the constant out:

\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

Apply u-substitution

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

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

Take the constant out:

\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

Apply the Power Rule

: \frac{1}{1 + ln ( 2 )} ln (2) v^{\frac{1 + l n ( 2 )}{l n ( 2 )}}

= \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 )}}

Substitute back

= \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 )}}

Simplify

\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 )}}

Add a constant to the solution

= \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

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What is the integral of 4^xe^{2x} ?
The integral of 4^xe^{2x} is 1/(2^{(1){ln(2)}}(1+ln(2))}*2^{\frac{2x+1+2ln(2)x-ln(2))/(ln(2))}+C