What is the integral of xye^x ?

The integral of xye^x is y(e^xx-e^x)+C

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integral of xye^x

\int x y e^{x} d x

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

Solution steps

\int x y e^{x} d x

Take the constant out:

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

= y \cdot \int e^{x} x d x

Apply Integration By Parts

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

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

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

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

Add a constant to the solution

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

\int (ln (x + 1 + x^{2}))^{'} d x

\frac{\partial}{\partial u} (u^{\frac{1}{3}} \cdot v^{\frac{2}{3}})

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

4 x^{3} - 5 7 x^{6} + π x

e^{- 4 t} cos (3 t)