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

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

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integral of 8e^{3x+e^{3x}}

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

\frac{8}{3} e^{e^{3 x}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 8 \cdot \int e^{u + 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

= 8 \cdot \frac{1}{3} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 8 \cdot \frac{1}{3} \cdot \int e^{v} d v

Use the common integral:

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

= 8 \cdot \frac{1}{3} e^{v}

Substitute back

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

Simplify

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

= \frac{8}{3} e^{e^{3 x}}

Add a constant to the solution

= \frac{8}{3} e^{e^{3 x}} + C

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What is the integral of 8e^{3x+e^{3x}} ?
The integral of 8e^{3x+e^{3x}} is 8/3 e^{e^{3x}}+C