What is the integral of ((e^{-t}-4))/(e^{-t)} ?

The integral of ((e^{-t}-4))/(e^{-t)} is t-4e^t+C

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integral of ((e^{-t}-4))/(e^{-t)}

integral

\frac{( e ^{- t} - 4 )}{e ^{- t}}

t - 4 e^{t} + C

Solution steps

\int \frac{e ^{- t} - 4}{e ^{- t}} d t

Simplify

\frac{e ^{- t} - 4}{e ^{- t}} : e^{t} (e^{- t} - 4)

= \int e^{t} (e^{- t} - 4) d t

Expand

e^{t} (e^{- t} - 4) : 1 - 4 e^{t}

= \int 1 - 4 e^{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int 1 d t - \int 4 e^{t} d t

\int 1 d t = t

\int 4 e^{t} d t = 4 e^{t}

= t - 4 e^{t}

Add a constant to the solution

= t - 4 e^{t} + C

What is the integral of ((e^{-t}-4))/(e^{-t)} ?
The integral of ((e^{-t}-4))/(e^{-t)} is t-4e^t+C