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 e^{u} \frac{1}{4} d u

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

x \to 2 lim (\frac{x - 5}{x - 2})

x \to 0 lim (\frac{z ^{x} - 1}{x})

x \to 0 lim (x + 4)

\frac{d}{d x} ((x^{- 2} + x^{- 3}) (x^{5} - 9 x^{2}))

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

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