What is the integral of (ye^x-ln(y^x)) ?

The integral of (ye^x-ln(y^x)) is -1/2 x^2ln(y)+ye^x+C

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integral of (ye^x-ln(y^x))

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

- \frac{1}{2} x^{2} ln (y) + y e^{x} + C

Solution steps

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

\int ln (y^{x}) d x = \frac{1}{2} x^{2} ln (y)

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

= - \frac{1}{2} x^{2} ln (y) + y e^{x}

Add a constant to the solution

= - \frac{1}{2} x^{2} ln (y) + y e^{x} + C

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What is the integral of (ye^x-ln(y^x)) ?
The integral of (ye^x-ln(y^x)) is -1/2 x^2ln(y)+ye^x+C