What is the integral of 6e^{2x+1} ?

The integral of 6e^{2x+1} is 3e^{2x+1}+C

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integral of 6e^{2x+1}

\int 6 e^{2 x + 1} d x

3 e^{2 x + 1} + C

Solution steps

\int 6 e^{2 x + 1} d x

Take the constant out:

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

= 6 \cdot \int e^{2 x + 1} d x

Apply u-substitution

= 6 \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

= 6 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{2} e^{u}

Substitute back

u = 2 x + 1

= 6 \cdot \frac{1}{2} e^{2 x + 1}

Simplify

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

= 3 e^{2 x + 1}

Add a constant to the solution

= 3 e^{2 x + 1} + C