What is the integral of 4x-10xe^{-0.2x} ?

The integral of 4x-10xe^{-0.2x} is 2x^2-250(-0.2e^{-0.2x}x-e^{-0.2x})+C

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integral of 4x-10xe^{-0.2x}

\int 4 x - 10 x e^{- 0.2 x} d x

2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

Solution steps

\int 4 x - 10 x e^{- 0.2 x} d x

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 x d x - \int 10 x e^{- 0.2 x} d x

\int 4 x d x = 2 x^{2}

\int 10 x e^{- 0.2 x} d x = 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x})

= 2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x})

Add a constant to the solution

= 2 x^{2} - 250 (- 0.2 e^{- 0.2 x} x - e^{- 0.2 x}) + C

What is the integral of 4x-10xe^{-0.2x} ?
The integral of 4x-10xe^{-0.2x} is 2x^2-250(-0.2e^{-0.2x}x-e^{-0.2x})+C