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

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

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integral of 5^{-5x}

\int 5^{- 5 x} d x

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

Solution steps

\int 5^{- 5 x} d x

Apply u-substitution

: \int - 5^{u - 1} d u

= \int - 5^{u - 1} d u

Take the constant out:

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

= - \int 5^{u - 1} d u

Simplify

5^{u - 1} : 5^{u} \cdot 5^{- 1}

= - \int 5^{u} \cdot 5^{- 1} d u

Take the constant out:

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

= - 5^{- 1} \cdot \int 5^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

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

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

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

Substitute back

u = - 5 x

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

Simplify

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

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

Add a constant to the solution

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

(x^{9} - 3 x)^{- 2}

\int (y + 1) d y

\frac{d y}{d x} - \frac{x y}{1 + x} = x

\frac{\partial}{\partial x} (ln (4 x^{2} + 9 y^{2} + 8))

y^{^{''}} + 2 y = 1

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