What is the integral of (t+1)e^{-t} ?

The integral of (t+1)e^{-t} is -e^{-t}t-2e^{-t}+C

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integral of (t+1)e^{-t}

\int (t + 1) e^{- t} d t

- e^{- t} t - 2 e^{- t} + C

Solution steps

\int (t + 1) e^{- t} d t

Apply Integration By Parts

: - e^{- t} (t + 1) - \int - e^{- t} d t

= - e^{- t} (t + 1) - \int - e^{- t} d t

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

= - e^{- t} (t + 1) - e^{- t}

Simplify

- e^{- t} (t + 1) - e^{- t} : - e^{- t} t - 2 e^{- t}

= - e^{- t} t - 2 e^{- t}

Add a constant to the solution

= - e^{- t} t - 2 e^{- t} + C

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What is the integral of (t+1)e^{-t} ?
The integral of (t+1)e^{-t} is -e^{-t}t-2e^{-t}+C