What is the integral of 5e^{-4x} ?

The integral of 5e^{-4x} is -5/4 e^{-4x}+C

Popular >

integral of 5e^{-4x}

\int 5 e^{- 4 x} d x

- \frac{5}{4} e^{- 4 x} + C

Solution steps

\int 5 e^{- 4 x} d x

Take the constant out:

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

= 5 \cdot \int e^{- 4 x} d x

Apply u-substitution

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

= 5 \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

= 5 (- \frac{1}{4} \cdot \int e^{u} d u)

Use the common integral:

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

= 5 (- \frac{1}{4} e^{u})

Substitute back

u = - 4 x

= 5 (- \frac{1}{4} e^{- 4 x})

Simplify

5 (- \frac{1}{4} e^{- 4 x}) : - \frac{5}{4} e^{- 4 x}

= - \frac{5}{4} e^{- 4 x}

Add a constant to the solution

= - \frac{5}{4} e^{- 4 x} + C

\frac{d}{d x} (arcsin (\frac{9}{x ^{2}}))

(\frac{4 x}{x ^{2} + 1})^{\frac{1}{4}}

f (x) = 4 x + 2

\int \frac{( y )}{( 1 + y ^{(2)} )} d y

y^{^{''}} + y^{^{'}} - 6 y = 0, y (0) = 2, y^{^{'}} (0) = - 1