What is the integral of 9xe^{-x} ?

The integral of 9xe^{-x} is 9(-e^{-x}x-e^{-x})+C

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integral of 9xe^{-x}

\int 9 x e^{- x} d x

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

Solution steps

\int 9 x e^{- x} d x

Take the constant out:

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

= 9 \cdot \int x e^{- x} d x

Apply Integration By Parts

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

= 9 (- e^{- x} x - \int - e^{- x} d x)

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

= 9 (- e^{- x} x - e^{- x})

Add a constant to the solution

= 9 (- e^{- x} x - e^{- x}) + C

\int_{0}^{1} x^{3} - 3 x^{2} + 2 x d x

y = x^{2} - 5 x + 5

\int x^{2} (6 - x) d x

2 y^{^{''}} - y^{^{'}} - y = 6 e^{- \frac{x}{2}}

\frac{d}{d x} (\frac{7}{x + 5})

What is the integral of 9xe^{-x} ?
The integral of 9xe^{-x} is 9(-e^{-x}x-e^{-x})+C