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

The integral of te^{-t}(2+t) is 4(-e^{-t}t-e^{-t})-e^{-t}t^2+C

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integral of te^{-t}(2+t)

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

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

Solution steps

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

Expand

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

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

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

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

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

Simplify

= 4 (- e^{- t} t - e^{- t}) - e^{- t} t^{2}

Add a constant to the solution

= 4 (- e^{- t} t - e^{- t}) - e^{- t} t^{2} + C

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What is the integral of te^{-t}(2+t) ?
The integral of te^{-t}(2+t) is 4(-e^{-t}t-e^{-t})-e^{-t}t^2+C