What is the integral of-(5^{-x})/(ln(5)) ?

The integral of-(5^{-x})/(ln(5)) is 1/(ln^2(5))*5^{-x}+C

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integral of-(5^{-x})/(ln(5))

\int - \frac{5 ^{- x}}{ln ( 5 )} d x

\frac{1}{ln ^{2} ( 5 )} \cdot 5^{- x} + C

Solution steps

\int - \frac{5 ^{- x}}{ln ( 5 )} d x

Take the constant out:

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

= - \frac{1}{ln ( 5 )} \cdot \int 5^{- x} d x

Apply u-substitution

: \int - 5^{u} d u

= - \frac{1}{ln ( 5 )} \cdot \int - 5^{u} d u

Take the constant out:

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

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

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

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

Substitute back

u = - x

= - \frac{1}{ln ( 5 )} (- \frac{5 ^{- x}}{ln ( 5 )})

Simplify

- \frac{1}{ln ( 5 )} (- \frac{5 ^{- x}}{ln ( 5 )}) : \frac{1}{ln ^{2} ( 5 )} \cdot 5^{- x}

= \frac{1}{ln ^{2} ( 5 )} \cdot 5^{- x}

Add a constant to the solution

= \frac{1}{ln ^{2} ( 5 )} \cdot 5^{- x} + C

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What is the integral of-(5^{-x})/(ln(5)) ?
The integral of-(5^{-x})/(ln(5)) is 1/(ln^2(5))*5^{-x}+C