What is the integral of 2te^{-t} ?

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

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integral of 2te^{-t}

\int 2 t e^{- t} d t

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

Solution steps

\int 2 t e^{- t} d t

Take the constant out:

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

= 2 \cdot \int t e^{- t} d t

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

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