What is the integral of 1/(\sqrt[5]{x^2)} ?

The integral of 1/(\sqrt[5]{x^2)} is 5/3 x^{3/5}+C

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integral of 1/(\sqrt[5]{x^2)}

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

Solution steps

assuming

x \geq 0

= \int \frac{1}{x ^{\frac{2}{5}}} d x

Apply the Power Rule

: \frac{5}{3} x^{\frac{3}{5}}

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

Add a constant to the solution

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

f (x) = (6 x + 1)^{2}

\frac{d}{d x} (ln (8 x))

2 \cdot x \cdot y \cdot y^{^{'}} = - (x^{2} + y^{2})

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

9 y^{7} + 4 x^{3} = 8 y + 5 x, (1, 1)

What is the integral of 1/(\sqrt[5]{x^2)} ?
The integral of 1/(\sqrt[5]{x^2)} is 5/3 x^{3/5}+C