What is the integral of 1/(5^{4x)} ?

The integral of 1/(5^{4x)} is -(625^{-x})/(4ln(5))+C

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integral of 1/(5^{4x)}

\int \frac{1}{5 ^{4 x}} d x

- \frac{62 5 ^{- x}}{4 ln ( 5 )} + C

Solution steps

\int \frac{1}{5 ^{4 x}} d x

Apply exponent rule:

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

\frac{1}{5 ^{4 x}} = 5^{- 4 x}

= \int 5^{- 4 x} d x

Simplify

5^{- 4 x} : 62 5^{- x}

= \int 62 5^{- x} d x

Apply u-substitution

: \int - 62 5^{u} d u

= \int - 62 5^{u} d u

Take the constant out:

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

= - \int 62 5^{u} d u

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

= - \frac{62 5 ^{u}}{ln ( 625 )}

Substitute back

u = - x

= - \frac{62 5 ^{- x}}{ln ( 625 )}

Simplify

ln (625) : 4 ln (5)

= - \frac{62 5 ^{- x}}{4 ln ( 5 )}

Add a constant to the solution

= - \frac{62 5 ^{- x}}{4 ln ( 5 )} + C

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{2} + 3 x - 1})

- 3 y^{^{'}} + \frac{2 y}{x} = \frac{( y ^{4} )}{x ^{4}}

\frac{d y}{d x} = \frac{x - y - 3}{x - 3}

\frac{1}{0.012 s ^{2} + 1.55 s + 50}

s \to 1 lim (\frac{( 2 s + 3 )}{s})

What is the integral of 1/(5^{4x)} ?
The integral of 1/(5^{4x)} is -(625^{-x})/(4ln(5))+C