What is the integral of 1/(sqrt(16-x)) ?

The integral of 1/(sqrt(16-x)) is -2sqrt(16-x)+C

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integral of 1/(sqrt(16-x))

\int \frac{1}{16 - x} d x

- 2 16 - x + C

Solution steps

\int \frac{1}{16 - x} d x

Apply u-substitution

: \int - \frac{1}{u} d u

= \int - \frac{1}{u} d u

Take the constant out:

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

= - \int \frac{1}{u} d u

Apply the Power Rule

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

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

Substitute back

u = 16 - x

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

Apply radical rule:

= - 2 16 - x

Add a constant to the solution

= - 2 16 - x + C

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What is the integral of 1/(sqrt(16-x)) ?
The integral of 1/(sqrt(16-x)) is -2sqrt(16-x)+C