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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int e^{v} d v

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d t} (2 e^{t} cos (t))

\frac{d P}{d t} + 2 tP = P + 6 t - 3

\int \frac{2 x + 5}{16 - 6 x - x ^{2}} d x

\int x^{3} x^{2} + 31 d x

x \to 0 lim (I n (x))

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