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

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

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

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

\frac{1}{3} e^{e^{3 x}} + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

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

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

Substitute back

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

Add a constant to the solution

= \frac{1}{3} e^{e^{3 x}} + C

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What is the integral of e^{e^{3x}+3x} ?
The integral of e^{e^{3x}+3x} is 1/3 e^{e^{3x}}+C