What is the integral of 3ye^{3z} ?

The integral of 3ye^{3z} is ye^{3z}+C

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integral of 3ye^{3z}

\int 3 y e^{3 z} d z

y e^{3 z} + C

Solution steps

\int 3 y e^{3 z} d z

Take the constant out:

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

= 3 y \cdot \int e^{3 z} d z

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= 3 y \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= 3 y \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 3 y \frac{1}{3} e^{u}

Substitute back

u = 3 z

= 3 y \frac{1}{3} e^{3 z}

Simplify

3 y \frac{1}{3} e^{3 z} : y e^{3 z}

= y e^{3 z}

Add a constant to the solution

= y e^{3 z} + C

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