What is the integral of 5e^{2x-1} ?

The integral of 5e^{2x-1} is 5/2 e^{2x-1}+C

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integral of 5e^{2x-1}

\int 5 e^{2 x - 1} d x

\frac{5}{2} e^{2 x - 1} + C

Solution steps

\int 5 e^{2 x - 1} d x

Take the constant out:

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

= 5 \cdot \int e^{2 x - 1} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 5 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x - 1

= 5 \cdot \frac{1}{2} e^{2 x - 1}

Simplify

5 \cdot \frac{1}{2} e^{2 x - 1} : \frac{5}{2} e^{2 x - 1}

= \frac{5}{2} e^{2 x - 1}

Add a constant to the solution

= \frac{5}{2} e^{2 x - 1} + C

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