What is the integral of-ge^{-ct} 1/c ?

The integral of-ge^{-ct} 1/c is e^{-ct} g/(c^2)+C

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integral of-ge^{-ct} 1/c

\int - g e^{- c t} \frac{1}{c} d t

e^{- c t} \frac{g}{c ^{2}} + C

Solution steps

\int - g e^{- c t} \frac{1}{c} d t

Take the constant out:

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

= - \frac{1}{c} g \cdot \int e^{- c t} d t

Simplify

- \frac{1}{c} g : - \frac{g}{c}

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

Apply u-substitution

= - \frac{g}{c} \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

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

Use the common integral:

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

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

Substitute back

u = - c t

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

Simplify

- \frac{g}{c} (- \frac{1}{c} e^{- c t}) : e^{- c t} \frac{g}{c ^{2}}

= e^{- c t} \frac{g}{c ^{2}}

Add a constant to the solution

= e^{- c t} \frac{g}{c ^{2}} + C

What is the integral of-ge^{-ct} 1/c ?
The integral of-ge^{-ct} 1/c is e^{-ct} g/(c^2)+C