What is the integral of 1/(sqrt(8-x)) ?

The integral of 1/(sqrt(8-x)) is -2sqrt(8-x)+C

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integral of 1/(sqrt(8-x))

\int \frac{1}{8 - x} d x

- 2 8 - x + C

Solution steps

\int \frac{1}{8 - 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 = 8 - x

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

Apply radical rule:

a^{\frac{1}{n}} = n a

= - 2 8 - x

Add a constant to the solution

= - 2 8 - x + C

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What is the integral of 1/(sqrt(8-x)) ?
The integral of 1/(sqrt(8-x)) is -2sqrt(8-x)+C