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

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

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

= - 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

\int e^{4 x} sin (3 x) d x

f (x) = \frac{cos ( x )}{x}

\int \frac{1}{a x ^{2} + b x} d x

\int \frac{sin ( 2 x )}{43 + cos ^{2} ( x )} d x

y^{^{'}} + 4 y = cos (x)

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