What is the integral of (8t+2)e^{-t} ?

The integral of (8t+2)e^{-t} is -8e^{-t}t-10e^{-t}+C

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integral of (8t+2)e^{-t}

\int (8 t + 2) e^{- t} d t

- 8 e^{- t} t - 10 e^{- t} + C

Solution steps

\int (8 t + 2) e^{- t} d t

Apply Integration By Parts

: - e^{- t} (8 t + 2) - \int - 8 e^{- t} d t

= - e^{- t} (8 t + 2) - \int - 8 e^{- t} d t

\int - 8 e^{- t} d t = 8 e^{- t}

= - e^{- t} (8 t + 2) - 8 e^{- t}

Simplify

- e^{- t} (8 t + 2) - 8 e^{- t} : - 8 e^{- t} t - 10 e^{- t}

= - 8 e^{- t} t - 10 e^{- t}

Add a constant to the solution

= - 8 e^{- t} t - 10 e^{- t} + C

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What is the integral of (8t+2)e^{-t} ?
The integral of (8t+2)e^{-t} is -8e^{-t}t-10e^{-t}+C