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

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

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

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

- \frac{1}{4} e^{- 4 x} + C

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

: \int - \frac{1}{4} e^{u} d u

= \int - \frac{1}{4} e^{u} d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= - \frac{1}{4} e^{u}

Substitute back

u = - 4 x

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

Add a constant to the solution

= - \frac{1}{4} e^{- 4 x} + C

\frac{d}{d x} (5 x^{6})

f (x) = \frac{2}{x - 1}, at x = 4

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

6 - 5 x

(\frac{( x + 11 )}{( x - 11 )})^{5}

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