What is the integral of (xye^{xz}) ?

The integral of (xye^{xz}) is ye^{xz}+C

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integral of (xye^{xz})

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

y e^{x z} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u}}{x} d u

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

Take the constant out:

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

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

Use the common integral:

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

= x y \frac{1}{x} e^{u}

Substitute back

u = x z

= x y \frac{1}{x} e^{x z}

Simplify

x y \frac{1}{x} e^{x z} : y e^{x z}

= y e^{x z}

Add a constant to the solution

= y e^{x z} + C

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