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

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

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integral of (yze^{xz})

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

y e^{z x} + C

Solution steps

\int yz e^{x z} d x

Take the constant out:

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

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

Apply u-substitution

= yz \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

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

Use the common integral:

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

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

Substitute back

u = z x

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

Simplify

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

= y e^{z x}

Add a constant to the solution

= y e^{z x} + C