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

\int x \cdot x^{2} + 9 d x

\frac{1}{( x ^{2} + 1 ) ^{2}}

\int 1 - (sin (x))^{2} d x

\int 2 x^{- \frac{3}{2}} d x

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