What is the integral of 3e^{6x} ?

The integral of 3e^{6x} is 1/2 e^{6x}+C

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integral of 3e^{6x}

\int 3 e^{6 x} d x

\frac{1}{2} e^{6 x} + C

Solution steps

\int 3 e^{6 x} d x

Take the constant out:

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

= 3 \cdot \int e^{6 x} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{6} \cdot \int e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{6} e^{u}

Substitute back

u = 6 x

= 3 \cdot \frac{1}{6} e^{6 x}

Simplify

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

= \frac{1}{2} e^{6 x}

Add a constant to the solution

= \frac{1}{2} e^{6 x} + C

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