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

The integral of 3\sqrt[5]{x^2} is 15/7 x^{7/5}+C

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

\int 3 5 x^{2} d x

\frac{15}{7} x^{\frac{7}{5}} + C

Solution steps

\int 3 5 x^{2} d x

Take the constant out:

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

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

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

assuming

x \geq 0

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

Apply the Power Rule

: \frac{5}{7} x^{\frac{7}{5}}

= 3 \cdot \frac{5}{7} x^{\frac{7}{5}}

Simplify

3 \cdot \frac{5}{7} x^{\frac{7}{5}} : \frac{15}{7} x^{\frac{7}{5}}

= \frac{15}{7} x^{\frac{7}{5}}

Add a constant to the solution

= \frac{15}{7} x^{\frac{7}{5}} + C

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What is the integral of 3\sqrt[5]{x^2} ?
The integral of 3\sqrt[5]{x^2} is 15/7 x^{7/5}+C