What is the integral of 5/(\sqrt[4]{x^2)} ?

The integral of 5/(\sqrt[4]{x^2)} is 10sqrt(x)+C

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integral of 5/(\sqrt[4]{x^2)}

10 x + 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

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

Apply the Power Rule

: 2 x^{\frac{1}{2}}

= 5 \cdot 2 x^{\frac{1}{2}}

Simplify

5 \cdot 2 x^{\frac{1}{2}} : 10 x

= 10 x

Add a constant to the solution

= 10 x + C

\frac{\partial}{\partial x} (ln (x^{2} + y^{2} + z^{2}))

\int (x) e^{x^{2} - 1} d x

x \to 2 lim (\frac{11 x}{4 - x ^{2}})

(3 x - 5) (x^{2} + 4)

\int \frac{7 x + 14}{x ^{2} + 5 x - 6} d x

What is the integral of 5/(\sqrt[4]{x^2)} ?
The integral of 5/(\sqrt[4]{x^2)} is 10sqrt(x)+C