What is the integral of e^{-4x-1} ?

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

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integral of e^{-4x-1}

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

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

Solution steps

\int e^{- 4 x - 1} 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 - 1

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

Add a constant to the solution

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

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