What is the integral of (16)/(sqrt(x)) ?

The integral of (16)/(sqrt(x)) is 32sqrt(x)+C

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integral of (16)/(sqrt(x))

\int \frac{16}{x} d x

32 x + C

Solution steps

\int \frac{16}{x} d x

Take the constant out:

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

= 16 \cdot \int \frac{1}{x} d x

Apply the Power Rule

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

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

Simplify

16 \cdot 2 x^{\frac{1}{2}} : 32 x

= 32 x

Add a constant to the solution

= 32 x + C

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What is the integral of (16)/(sqrt(x)) ?
The integral of (16)/(sqrt(x)) is 32sqrt(x)+C