What is the integral of (e^{xy}+xye^{xy}) ?

The integral of (e^{xy}+xye^{xy}) is 1/y e^{yx}+e^{yx}x-(e^{yx})/y+C

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integral of (e^{xy}+xye^{xy})

\int (e^{x y} + x y e^{x y}) d x

\frac{1}{y} e^{y x} + e^{y x} x - \frac{e ^{y x}}{y} + C

Solution steps

\int e^{x y} + x y e^{x y} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{y x} d x = \frac{1}{y} e^{y x}

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

= \frac{1}{y} e^{y x} + e^{y x} x - \frac{e ^{y x}}{y}

Add a constant to the solution

= \frac{1}{y} e^{y x} + e^{y x} x - \frac{e ^{y x}}{y} + C

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What is the integral of (e^{xy}+xye^{xy}) ?
The integral of (e^{xy}+xye^{xy}) is 1/y e^{yx}+e^{yx}x-(e^{yx})/y+C