What is the integral of 6e^{3x+7} ?

The integral of 6e^{3x+7} is 2e^{3x+7}+C

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integral of 6e^{3x+7}

\int 6 e^{3 x + 7} d x

2 e^{3 x + 7} + C

Solution steps

\int 6 e^{3 x + 7} d x

Take the constant out:

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

= 6 \cdot \int e^{3 x + 7} d x

Apply u-substitution

= 6 \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

= 6 \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

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

= 6 \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x + 7

= 6 \cdot \frac{1}{3} e^{3 x + 7}

Simplify

6 \cdot \frac{1}{3} e^{3 x + 7} : 2 e^{3 x + 7}

= 2 e^{3 x + 7}

Add a constant to the solution

= 2 e^{3 x + 7} + C