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

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

Popular >

integral of-e^{4x}

\int - e^{4 x} d x

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

Solution steps

\int - e^{4 x} d x

Take the constant out:

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

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

Apply u-substitution

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

= - \int e^{u} \frac{1}{4} 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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{n}{csc ( \frac{1}{n} )}

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

\frac{\partial}{\partial x} (y ln (x^{3} + y^{4} + 2))

\frac{d}{d x} (e^{x z (x)})

f (x) = x^{2} + 169 x + 820, at x = 2