What is the integral of-9x*e^x ?

The integral of-9x*e^x is -9(e^xx-e^x)+C

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integral of-9x*e^x

\int - 9 x \cdot 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

n = 0 \sum \infty \frac{n}{( n ^{2} + 1 )}

\int (x^{3} - 2 x) d x

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

y = \frac{6}{x}, at x = 2

\frac{d}{d x} (2 x (1 - x))