What is the integral of 1/(x^{4/3)} ?

The integral of 1/(x^{4/3)} is -3/(x^{1/3)}+C

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integral of 1/(x^{4/3)}

\int \frac{1}{x ^{\frac{4}{3}}} d x

- \frac{3}{x ^{\frac{1}{3}}} + C

Solution steps

\int \frac{1}{x ^{\frac{4}{3}}} d x

Apply the Power Rule

: - \frac{3}{x ^{\frac{1}{3}}}

= - \frac{3}{x ^{\frac{1}{3}}}

Add a constant to the solution

= - \frac{3}{x ^{\frac{1}{3}}} + C

\frac{\partial}{\partial y} (x^{2} - y^{3} - x^{2} y + y)

\int_{1}^{R} \frac{ln ( 4 x )}{x ^{2}} d x

\int \frac{e}{π} d x

\frac{\partial}{\partial x} (x^{2} e^{4 x y})

\frac{e ^{x}}{x ^{n}}

What is the integral of 1/(x^{4/3)} ?
The integral of 1/(x^{4/3)} is -3/(x^{1/3)}+C