What is the integral of 9xe^x ?

The integral of 9xe^x is 9(e^xx-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

y^{^{''}} + 4 \cdot y = cos (x)

\frac{d}{d x} (\frac{x}{x ^{2} + 225})

\int_{1}^{4} x^{2} - 5 x + 4 d x

x^{\frac{3}{2}} + x^{\frac{1}{2}}

x \to 1 lim (\frac{3 - x}{3 + x} e^{- x} cos (π x))