What is the integral of ge^{-cx} ?

The integral of ge^{-cx} is -ge^{-cx} 1/c+C

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integral of ge^{-cx}

\int g e^{- c x} d x

- g e^{- c x} \frac{1}{c} + C

Solution steps

\int g e^{- c x} d x

Take the constant out:

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

= g \cdot \int e^{- c x} d x

Apply u-substitution

: \int - \frac{e ^{u}}{c} d u

= g \cdot \int - \frac{e ^{u}}{c} d u

Take the constant out:

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

= g (- \frac{1}{c} \cdot \int e^{u} d u)

Use the common integral:

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

= g (- \frac{1}{c} e^{u})

Substitute back

u = - c x

= g (- \frac{1}{c} e^{- c x})

Simplify

= - g e^{- c x} \frac{1}{c}

Add a constant to the solution

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

f (x) = ln (x^{2} - 2 x + 1), at x = 2

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

\int e^{- 0.06 x} d x

\frac{\partial}{\partial y} (sin (x - y) + (x y))

\frac{\partial}{\partial x} (- 4 x^{2} y - 4 x)