What is the integral of (16)/(e^{4x)} ?

The integral of (16)/(e^{4x)} is -4e^{-4x}+C

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integral of (16)/(e^{4x)}

\int \frac{16}{e ^{4 x}} d x

- 4 e^{- 4 x} + C

Solution steps

\int \frac{16}{e ^{4 x}} d x

Take the constant out:

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

= 16 \cdot \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}

= 16 \cdot \int e^{- 4 x} d x

Apply u-substitution

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

= 16 \cdot \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

= 16 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

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

= 16 (- \frac{1}{4} e^{u})

Substitute back

u = - 4 x

= 16 (- \frac{1}{4} e^{- 4 x})

Simplify

16 (- \frac{1}{4} e^{- 4 x}) : - 4 e^{- 4 x}

= - 4 e^{- 4 x}

Add a constant to the solution

= - 4 e^{- 4 x} + C

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What is the integral of (16)/(e^{4x)} ?
The integral of (16)/(e^{4x)} is -4e^{-4x}+C