What is the integral of 8xe^{-x} ?

The integral of 8xe^{-x} is 8(-e^{-x}x-e^{-x})+C

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integral of 8xe^{-x}

\int 8 x e^{- x} d x

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

Solution steps

\int 8 x e^{- x} d x

Take the constant out:

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

= 8 \cdot \int x e^{- x} d x

Apply Integration By Parts

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

= 8 (- e^{- x} x - \int - e^{- x} d x)

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

= 8 (- e^{- x} x - e^{- x})

Add a constant to the solution

= 8 (- e^{- x} x - e^{- x}) + C

y = 2 - e^{x}

f (x) = 5 - 5 x^{2}, (- 2, - 15)

f (x) = 5^{1 - x}

\int \frac{x ^{2} + 1}{x ^{3} + x ^{2} - 2 x} d x

\int \frac{e ^{2 x}}{e ^{2 x}} d x

What is the integral of 8xe^{-x} ?
The integral of 8xe^{-x} is 8(-e^{-x}x-e^{-x})+C