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

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

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integral of 4^xe^{-x}

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

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

Solution steps

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

Apply u-substitution

: \int - 4^{- l n (u)} d u

= \int - 4^{- l n (u)} d u

Take the constant out:

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

= - \int 4^{- l n (u)} d u

Apply Integration By Parts

: 4^{- l n (u)} u - \int - \frac{2 ln ( 2 )}{u ^{2 l n (2)}} d u

= - (4^{- l n (u)} u - \int - \frac{2 ln ( 2 )}{u ^{2 l n (2)}} d u)

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

= - (4^{- l n (u)} u - (- \frac{2}{- 2 ln ( 2 ) + 1} ln (2) u^{- 2 l n (2) + 1}))

Substitute back

u = e^{- x}

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

Simplify

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

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

Add a constant to the solution

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

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