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

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

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integral of 4xe^{-x}

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

4 (- e^{- x} x - e^{- x}) + C

Solution steps

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

Take the constant out:

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

= 4 \cdot \int x e^{- x} d x

Apply Integration By Parts

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

= 4 (- e^{- x} x - \int - e^{- x} d x)

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

= 4 (- e^{- x} x - e^{- x})

Add a constant to the solution

= 4 (- e^{- x} x - e^{- x}) + C

y = 4 x

\frac{\partial}{\partial x} (x^{3} y^{2} - 4 y x^{2} + 3 y x - 4)

f (x) = \frac{x}{x ^{2} - 9}

y = x^{2} + 5 x + 6, at x = 7

\int 144 - 9 x^{2} d x

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