What is the integral of \sqrt[3]{e^{4x}+2e^{4x}} ?

The integral of \sqrt[3]{e^{4x}+2e^{4x}} is \sqrt[3]{3} 3/4 e^{(4x)/3}+C

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integral of \sqrt[3]{e^{4x}+2e^{4x}}

Solution steps

Simplify

Take the constant out:

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

Apply u-substitution

: \int e^{u} \frac{3}{4} d u

Take the constant out:

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

Use the common integral:

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

Substitute back

u = \frac{4 x}{3}

Add a constant to the solution

\frac{d}{d x} (x^{3} - 12 x y + 8 y^{3})

\int (x - 1) (x + 2) d x

f (x) = 5 x^{\frac{4}{5}} - 2 x^{\frac{2}{7}}

\frac{1}{x + 5}

\int t (2 t - 3 t) d t

What is the integral of \sqrt[3]{e^{4x}+2e^{4x}} ?
The integral of \sqrt[3]{e^{4x}+2e^{4x}} is \sqrt[3]{3} 3/4 e^{(4x)/3}+C