What is the integral of 5e^{3x+e^{3x}} ?

The integral of 5e^{3x+e^{3x}} is 5/3 e^{e^{3x}}+C

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integral of 5e^{3x+e^{3x}}

\int 5 e^{3 x + e^{3 x}} d x

\frac{5}{3} e^{e^{3 x}} + C

Solution steps

\int 5 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

= 5 \cdot \int e^{3 x + e^{3 x}} d x

Apply u-substitution

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

= 5 \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

= 5 \cdot \frac{1}{3} \cdot \int e^{u + e^{u}} d u

Apply u-substitution

: \int e^{v} d v

= 5 \cdot \frac{1}{3} \cdot \int e^{v} d v

Use the common integral:

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

= 5 \cdot \frac{1}{3} e^{v}

Substitute back

= 5 \cdot \frac{1}{3} e^{e^{3 x}}

Simplify

5 \cdot \frac{1}{3} e^{e^{3 x}} : \frac{5}{3} e^{e^{3 x}}

= \frac{5}{3} e^{e^{3 x}}

Add a constant to the solution

= \frac{5}{3} e^{e^{3 x}} + C

f (x) = sec^{3} (x)

\int (t^{2} + 2) d t

\int \frac{e ^{x}}{e ^{x} ( e ^{x} - 3 )} d x

\frac{( 2 x ) ^{3}}{5}

\frac{d y}{d x} (y + 4) = 2 x y + 4 y

What is the integral of 5e^{3x+e^{3x}} ?
The integral of 5e^{3x+e^{3x}} is 5/3 e^{e^{3x}}+C