What is the integral of e^{(4x)/9}(4/9) ?

The integral of e^{(4x)/9}(4/9) is e^{(4x)/9}+C

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integral of e^{(4x)/9}(4/9)

\int e^{\frac{4 x}{9}} (\frac{4}{9}) d x

e^{\frac{4 x}{9}} + C

Solution steps

\int e^{\frac{4 x}{9}} \frac{4}{9} d x

Take the constant out:

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

= \frac{4}{9} \cdot \int e^{\frac{4 x}{9}} d x

Apply u-substitution

= \frac{4}{9} \cdot \int e^{u} \frac{9}{4} d u

Take the constant out:

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

= \frac{4}{9} \cdot \frac{9}{4} \cdot \int e^{u} d u

Use the common integral:

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

= \frac{4}{9} \cdot \frac{9}{4} e^{u}

Substitute back

u = \frac{4 x}{9}

= \frac{4}{9} \cdot \frac{9}{4} e^{\frac{4 x}{9}}

Simplify

\frac{4}{9} \cdot \frac{9}{4} e^{\frac{4 x}{9}} : e^{\frac{4 x}{9}}

= e^{\frac{4 x}{9}}

Add a constant to the solution

= e^{\frac{4 x}{9}} + C

What is the integral of e^{(4x)/9}(4/9) ?
The integral of e^{(4x)/9}(4/9) is e^{(4x)/9}+C