What is the integral of 2e^{3x-5} ?

The integral of 2e^{3x-5} is 2/3 e^{3x-5}+C

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integral of 2e^{3x-5}

\int 2 e^{3 x - 5} d x

\frac{2}{3} e^{3 x - 5} + C

Solution steps

\int 2 e^{3 x - 5} d x

Take the constant out:

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

= 2 \cdot \int e^{3 x - 5} d x

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

= 2 \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 2 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x - 5

= 2 \cdot \frac{1}{3} e^{3 x - 5}

Simplify

2 \cdot \frac{1}{3} e^{3 x - 5} : \frac{2}{3} e^{3 x - 5}

= \frac{2}{3} e^{3 x - 5}

Add a constant to the solution

= \frac{2}{3} e^{3 x - 5} + C

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