What is the integral of ze^{yz} ?

The integral of ze^{yz} is e^{zy}+C

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integral of ze^{yz}

\int z e^{yz} d y

e^{zy} + C

Solution steps

\int z e^{yz} d y

Take the constant out:

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

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

Apply u-substitution

= 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

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

Use the common integral:

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

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

Substitute back

u = zy

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

Simplify

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

= e^{zy}

Add a constant to the solution

= e^{zy} + C