What is the integral of e^{9x-5} ?

The integral of e^{9x-5} is 1/9 e^{9x-5}+C

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integral of e^{9x-5}

\int e^{9 x - 5} d x

\frac{1}{9} e^{9 x - 5} + C

Solution steps

\int e^{9 x - 5} d x

Apply u-substitution

= \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

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

Use the common integral:

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

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

Substitute back

u = 9 x - 5

= \frac{1}{9} e^{9 x - 5}

Add a constant to the solution

= \frac{1}{9} e^{9 x - 5} + C

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