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

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

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integral of 9e^{2x}

\int 9 e^{2 x} d x

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

Solution steps

\int 9 e^{2 x} d x

Take the constant out:

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

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

Apply u-substitution

: \int e^{u} \frac{1}{2} d u

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = 2 x

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

Simplify

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

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

Add a constant to the solution

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

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