What is the integral of 9e^{6y} ?

The integral of 9e^{6y} is 3/2 e^{6y}+C

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integral of 9e^{6y}

\int 9 e^{6 y} d y

\frac{3}{2} e^{6 y} + C

Solution steps

\int 9 e^{6 y} d y

Take the constant out:

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

= 9 \cdot \int e^{6 y} d y

Apply u-substitution

: \int e^{u} \frac{1}{6} d u

= 9 \cdot \int e^{u} \frac{1}{6} d u

Take the constant out:

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

= 9 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{6} e^{u}

Substitute back

u = 6 y

= 9 \cdot \frac{1}{6} e^{6 y}

Simplify

9 \cdot \frac{1}{6} e^{6 y} : \frac{3}{2} e^{6 y}

= \frac{3}{2} e^{6 y}

Add a constant to the solution

= \frac{3}{2} e^{6 y} + C

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