What is the integral of (2e^{9x}) ?

The integral of (2e^{9x}) is 2/9 e^{9x}+C

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integral of (2e^{9x})

\int (2 e^{9 x}) d x

\frac{2}{9} e^{9 x} + C

Solution steps

\int 2 e^{9 x} d x

Take the constant out:

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

= 2 \cdot \int e^{9 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{9} d u

= 2 \cdot \int e^{u} \frac{1}{9} d u

Take the constant out:

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

= 2 \cdot \frac{1}{9} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{9} e^{u}

Substitute back

u = 9 x

= 2 \cdot \frac{1}{9} e^{9 x}

Simplify

2 \cdot \frac{1}{9} e^{9 x} : \frac{2}{9} e^{9 x}

= \frac{2}{9} e^{9 x}

Add a constant to the solution

= \frac{2}{9} e^{9 x} + C

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