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

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

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integral of e^{-t}2te^{2t}

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

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

Solution steps

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

Take the constant out:

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

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

Simplify

e^{- t} t e^{2 t} : e^{t} t

= 2 \cdot \int e^{t} 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 e^{-t}2te^{2t} ?
The integral of e^{-t}2te^{2t} is 2(e^tt-e^t)+C