What is the integral of 4/(\sqrt[5]{x^3)} ?

The integral of 4/(\sqrt[5]{x^3)} is 10x^{2/5}+C

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integral of 4/(\sqrt[5]{x^3)}

10 x^{\frac{2}{5}} + C

Solution steps

Take the constant out:

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

Apply exponent rule:

(a^{n})^{m} = a^{n \cdot m}

= 4 \cdot \int \frac{1}{x ^{\frac{3}{5}}} d x

Apply the Power Rule

: \frac{5}{2} x^{\frac{2}{5}}

= 4 \cdot \frac{5}{2} x^{\frac{2}{5}}

Simplify

4 \cdot \frac{5}{2} x^{\frac{2}{5}} : 10 x^{\frac{2}{5}}

= 10 x^{\frac{2}{5}}

Add a constant to the solution

= 10 x^{\frac{2}{5}} + C

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

x \to 0 lim (\frac{1}{csc ( x )})

\frac{d}{d x} (\frac{2 x + 1}{3 x + 4})

\frac{\partial}{\partial y} (\frac{x ^{2}}{y})

\int sin (x) + cos (x) d x

What is the integral of 4/(\sqrt[5]{x^3)} ?
The integral of 4/(\sqrt[5]{x^3)} is 10x^{2/5}+C