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

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

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integral of (4t-2)(e^{t^2-t})

\int (4 t - 2) (e^{t^{2} - t}) d t

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

Solution steps

\int (4 t - 2) e^{t^{2} - t} d t

Expand

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

= \int 4 e^{t^{2} - t} t - 2 e^{t^{2} - t} 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 4 e^{t^{2} - t} t d t - \int 2 e^{t^{2} - t} d t

Simplify

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

Add a constant to the solution

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

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