What is the integral of t^3-(e^{-t}-4)/(e^{-t)} ?

The integral of t^3-(e^{-t}-4)/(e^{-t)} is (t^4)/4-t+4e^t+C

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integral of t^3-(e^{-t}-4)/(e^{-t)}

\int t^{3} - \frac{e ^{- t} - 4}{e ^{- t}} d t

\frac{t ^{4}}{4} - t + 4 e^{t} + C

Solution steps

\int t^{3} - \frac{e ^{- t} - 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 t^{3} d t - \int \frac{e ^{- t} - 4}{e ^{- t}} d t

\int t^{3} d t = \frac{t ^{4}}{4}

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

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

Simplify

= \frac{t ^{4}}{4} - t + 4 e^{t}

Add a constant to the solution

= \frac{t ^{4}}{4} - t + 4 e^{t} + C

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What is the integral of t^3-(e^{-t}-4)/(e^{-t)} ?
The integral of t^3-(e^{-t}-4)/(e^{-t)} is (t^4)/4-t+4e^t+C