What is the integral of 8x^{-3/2} ?

The integral of 8x^{-3/2} is -(16)/(x^{1/2)}+C

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integral of 8x^{-3/2}

\int 8 x^{- \frac{3}{2}} d x

- \frac{16}{x ^{\frac{1}{2}}} + C

Solution steps

\int 8 x^{- \frac{3}{2}} d x

Take the constant out:

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

= 8 \cdot \int x^{- \frac{3}{2}} d x

Apply the Power Rule

= 8 (- \frac{2}{x ^{\frac{1}{2}}})

Simplify

8 (- \frac{2}{x ^{\frac{1}{2}}}) : - \frac{16}{x ^{\frac{1}{2}}}

= - \frac{16}{x ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{16}{x ^{\frac{1}{2}}} + C

What is the integral of 8x^{-3/2} ?
The integral of 8x^{-3/2} is -(16)/(x^{1/2)}+C