What is the integral of 16e^{4x} ?

The integral of 16e^{4x} is 4e^{4x}+C

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integral of 16e^{4x}

\int 16 e^{4 x} d x

4 e^{4 x} + C

Solution steps

\int 16 e^{4 x} d x

Take the constant out:

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

= 16 \cdot \int e^{4 x} d x

Apply u-substitution

= 16 \cdot \int e^{u} \frac{1}{4} d u

Take the constant out:

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

= 16 \cdot \frac{1}{4} \cdot \int e^{u} d u

Use the common integral:

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

= 16 \cdot \frac{1}{4} e^{u}

Substitute back

u = 4 x

= 16 \cdot \frac{1}{4} e^{4 x}

Simplify

16 \cdot \frac{1}{4} e^{4 x} : 4 e^{4 x}

= 4 e^{4 x}

Add a constant to the solution

= 4 e^{4 x} + C