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

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

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integral of 4e^{-x/4}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

= 4 (- 4 e^{u})

Substitute back

u = - \frac{x}{4}

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

Simplify

= - 16 e^{- \frac{x}{4}}

Add a constant to the solution

= - 16 e^{- \frac{x}{4}} + C

f (x) = - 2 sin (x)

\frac{5 + 5 ^{2} + 4 x ^{2}}{2}

e^{x} + 17

\frac{d}{d x} (ln (\frac{cos ( x )}{sin ( x )}))

- e^{t} + \frac{1}{2} sin (t) + \frac{1}{2} cos (t)