What is the integral of 23e^{39x} ?

The integral of 23e^{39x} is 23/39 e^{39x}+C

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integral of 23e^{39x}

\int 23 e^{39 x} d x

\frac{23}{39} e^{39 x} + C

Solution steps

\int 23 e^{39 x} d x

Take the constant out:

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

= 23 \cdot \int e^{39 x} d x

Apply u-substitution

: \int e^{u} \frac{1}{39} d u

= 23 \cdot \int e^{u} \frac{1}{39} d u

Take the constant out:

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

= 23 \cdot \frac{1}{39} \cdot \int e^{u} d u

Use the common integral:

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

= 23 \cdot \frac{1}{39} e^{u}

Substitute back

u = 39 x

= 23 \cdot \frac{1}{39} e^{39 x}

Simplify

23 \cdot \frac{1}{39} e^{39 x} : \frac{23}{39} e^{39 x}

= \frac{23}{39} e^{39 x}

Add a constant to the solution

= \frac{23}{39} e^{39 x} + C

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What is the integral of 23e^{39x} ?
The integral of 23e^{39x} is 23/39 e^{39x}+C