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

: \int \frac{e ^{u}}{2} d u

= 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

\frac{d}{d x} (- 4 x^{\frac{1}{4}} + 3 x^{- 6})

\frac{\partial}{\partial x} (e^{x - 3 y})

\int_{0}^{2} e^{- 2 x} d x

f (x) = 8 x e^{x}

\frac{d y}{d t} = 0.75 y - 0.00003 y^{2}

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