What is the integral of (ye^{xy}-1/y) ?

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

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integral of (ye^{xy}-1/y)

\int (y e^{x y} - \frac{1}{y}) d x

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

Solution steps

\int y e^{x y} - \frac{1}{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 \frac{1}{y} d x + \int y e^{y x} d x

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

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

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

Add a constant to the solution

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

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