What is the integral of 5^xe^{-x} ?

The integral of 5^xe^{-x} is -5^xe^{-x}-1/(-ln(5)+1)ln(5)e^{-(-ln(5)+1)x}+C

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integral of 5^xe^{-x}

\int 5^{x} e^{- x} d x

- 5^{x} e^{- x} - \frac{1}{- ln ( 5 ) + 1} ln (5) e^{- (- l n (5) + 1) x} + C

Solution steps

\int 5^{x} e^{- x} d x

Apply u-substitution

: \int - 5^{- l n (u)} d u

= \int - 5^{- l n (u)} d u

Take the constant out:

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

= - \int 5^{- l n (u)} d u

Apply Integration By Parts

: 5^{- l n (u)} u - \int - \frac{ln ( 5 )}{u ^{l n (5)}} d u

= - (5^{- l n (u)} u - \int - \frac{ln ( 5 )}{u ^{l n (5)}} d u)

\int - \frac{ln ( 5 )}{u ^{l n (5)}} d u = - \frac{1}{- ln ( 5 ) + 1} ln (5) u^{- l n (5) + 1}

= - (5^{- l n (u)} u - (- \frac{1}{- ln ( 5 ) + 1} ln (5) u^{- l n (5) + 1}))

Substitute back

u = e^{- x}

= - (5^{- l n (e^{- x})} e^{- x} - (- \frac{1}{- ln ( 5 ) + 1} ln (5) (e^{- x})^{- l n (5) + 1}))

Simplify

- (5^{- l n (e^{- x})} e^{- x} - (- \frac{1}{- ln ( 5 ) + 1} ln (5) (e^{- x})^{- l n (5) + 1})) : - 5^{x} e^{- x} - \frac{1}{- ln ( 5 ) + 1} ln (5) e^{- (- l n (5) + 1) x}

= - 5^{x} e^{- x} - \frac{1}{- ln ( 5 ) + 1} ln (5) e^{- (- l n (5) + 1) x}

Add a constant to the solution

= - 5^{x} e^{- x} - \frac{1}{- ln ( 5 ) + 1} ln (5) e^{- (- l n (5) + 1) x} + C

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What is the integral of 5^xe^{-x} ?
The integral of 5^xe^{-x} is -5^xe^{-x}-1/(-ln(5)+1)ln(5)e^{-(-ln(5)+1)x}+C