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

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

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integral of 3/(\sqrt[5]{x^2)}

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

Solution steps

Take the constant out:

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

assuming

x \geq 0

= 3 \cdot \int \frac{1}{x ^{\frac{2}{5}}} d x

Apply the Power Rule

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{3 ( 1 - 4 x )}{58 ( x ^{2} + 9 )}

sin (ln (cos (t)))

\int x^{2} 6 x + 6 d x

f (x) = x^{3} + x + 5, at x = - 2

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