What is the integral of 9e^{2x+e^{2x}} ?

The integral of 9e^{2x+e^{2x}} is 9/2 e^{e^{2x}}+C

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integral of 9e^{2x+e^{2x}}

\int 9 e^{2 x + e^{2 x}} d x

\frac{9}{2} e^{e^{2 x}} + C

Solution steps

\int 9 e^{2 x + e^{2 x}} d x

Take the constant out:

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

= 9 \cdot \int e^{2 x + e^{2 x}} d x

Apply u-substitution

= 9 \cdot \int \frac{e ^{u}}{2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 9 \cdot \frac{1}{2} e^{u}

Substitute back

u = e^{2 x}

= 9 \cdot \frac{1}{2} e^{e^{2 x}}

Simplify

9 \cdot \frac{1}{2} e^{e^{2 x}} : \frac{9}{2} e^{e^{2 x}}

= \frac{9}{2} e^{e^{2 x}}

Add a constant to the solution

= \frac{9}{2} e^{e^{2 x}} + C

What is the integral of 9e^{2x+e^{2x}} ?
The integral of 9e^{2x+e^{2x}} is 9/2 e^{e^{2x}}+C