What is the integral of xze^{yz} ?

The integral of xze^{yz} is xe^{zy}+C

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integral of xze^{yz}

\int x z e^{yz} d y

x e^{zy} + C

Solution steps

\int x z e^{yz} d y

Take the constant out:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = zy

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

Simplify

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

= x e^{zy}

Add a constant to the solution

= x e^{zy} + C

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