What is the integral of 1/(2x^{3/2)} ?

The integral of 1/(2x^{3/2)} is -1/(x^{1/2)}+C

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integral of 1/(2x^{3/2)}

\int \frac{1}{2 x ^{\frac{3}{2}}} d x

- \frac{1}{x ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{2 x ^{\frac{3}{2}}} d x

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} (- \frac{2}{x ^{\frac{1}{2}}})

Simplify

\frac{1}{2} (- \frac{2}{x ^{\frac{1}{2}}}) : - \frac{1}{x ^{\frac{1}{2}}}

= - \frac{1}{x ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{1}{x ^{\frac{1}{2}}} + C

What is the integral of 1/(2x^{3/2)} ?
The integral of 1/(2x^{3/2)} is -1/(x^{1/2)}+C