What is the integral of 8e^{2x+3} ?

The integral of 8e^{2x+3} is 4e^{2x+3}+C

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integral of 8e^{2x+3}

\int 8 e^{2 x + 3} d x

4 e^{2 x + 3} + C

Solution steps

\int 8 e^{2 x + 3} d x

Take the constant out:

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

= 8 \cdot \int e^{2 x + 3} d x

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 8 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x + 3

= 8 \cdot \frac{1}{2} e^{2 x + 3}

Simplify

8 \cdot \frac{1}{2} e^{2 x + 3} : 4 e^{2 x + 3}

= 4 e^{2 x + 3}

Add a constant to the solution

= 4 e^{2 x + 3} + C

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