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

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

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

\frac{15}{2} 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}

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

Apply the Power Rule

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

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

Simplify

3 \cdot \frac{5}{2} x^{\frac{2}{5}} : \frac{15}{2} x^{\frac{2}{5}}

= \frac{15}{2} x^{\frac{2}{5}}

Add a constant to the solution

= \frac{15}{2} x^{\frac{2}{5}} + C

\frac{d}{d x} ((3 x^{2} - 2 x)^{2})

x^{2} \frac{d y}{d x} + y = 0

\frac{d}{d x} (sin^{2} (6 x))

\frac{d}{d x} (1 + x - 1)

(x^{3} - 25 x^{2})^{^{'}}

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