What is the integral of 3e^{2x+e^{2x}} ?

The integral of 3e^{2x+e^{2x}} is 3/2 e^{e^{2x}}+C

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integral of 3e^{2x+e^{2x}}

\int 3 e^{2 x + e^{2 x}} d x

\frac{3}{2} e^{e^{2 x}} + C

Solution steps

\int 3 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

= 3 \cdot \int e^{2 x + e^{2 x}} d x

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{2} \cdot \int e^{u} d u

Use the common integral:

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

= 3 \cdot \frac{1}{2} e^{u}

Substitute back

u = e^{2 x}

= 3 \cdot \frac{1}{2} e^{e^{2 x}}

Simplify

3 \cdot \frac{1}{2} e^{e^{2 x}} : \frac{3}{2} e^{e^{2 x}}

= \frac{3}{2} e^{e^{2 x}}

Add a constant to the solution

= \frac{3}{2} e^{e^{2 x}} + C

\frac{6}{x}

\int (2 x^{4} - 1) (5 x^{3} + 6 x) d x

f (- 3) = - 2 x^{2} + 4 x - \frac{7}{x ^{2}}

\frac{d y}{d x} = y (x y^{4} - 1)

f (x) = - \frac{4}{x}

What is the integral of 3e^{2x+e^{2x}} ?
The integral of 3e^{2x+e^{2x}} is 3/2 e^{e^{2x}}+C