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

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

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integral of 2e^{2e^x+x}

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

e^{2 e^{x}} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{2 u} d u

= 2 \cdot \int e^{2 u} d u

Apply u-substitution

: \int e^{v} \frac{1}{2} d v

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

Take the constant out:

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

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

Use the common integral:

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

= 2 \cdot \frac{1}{2} e^{v}

Substitute back

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

Simplify

2 \cdot \frac{1}{2} e^{2 e^{x}} : e^{2 e^{x}}

= e^{2 e^{x}}

Add a constant to the solution

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

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