What is the integral of e^{-2kx} ?

The integral of e^{-2kx} is -1/(2k)e^{-2kx}+C

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integral of e^{-2kx}

\int e^{- 2 k x} d x

- \frac{1}{2 k} e^{- 2 k x} + C

Solution steps

\int e^{- 2 k x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{2 k} d u

= \int - \frac{e ^{u}}{2 k} d u

Take the constant out:

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

= - \frac{1}{2 k} \cdot \int e^{u} d u

Use the common integral:

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

= - \frac{1}{2 k} e^{u}

Substitute back

u = - 2 k x

= - \frac{1}{2 k} e^{- 2 k x}

Add a constant to the solution

= - \frac{1}{2 k} e^{- 2 k x} + C

y = 7 + 9 x

\frac{d}{d x} (5 sec (x) - 10 cos (x))

(x^{2} + 3) y^{^{'}} = x y

(x^{3} - 9 x)^{12}

x (\frac{d y}{d x}) - y = 3 x^{4}

What is the integral of e^{-2kx} ?
The integral of e^{-2kx} is -1/(2k)e^{-2kx}+C