What is the integral of 1/(sqrt(6-x)) ?

The integral of 1/(sqrt(6-x)) is -2sqrt(6-x)+C

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integral of 1/(sqrt(6-x))

\int \frac{1}{6 - x} d x

- 2 6 - x + C

Solution steps

\int \frac{1}{6 - 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 = 6 - x

= - 2 (6 - x)^{\frac{1}{2}}

Apply radical rule:

= - 2 6 - x

Add a constant to the solution

= - 2 6 - x + C

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\int \frac{1}{x ^{2} - x + 1} d x

\frac{\partial}{\partial x} (2 (x^{2} + 4)^{- \frac{1}{2}})

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What is the integral of 1/(sqrt(6-x)) ?
The integral of 1/(sqrt(6-x)) is -2sqrt(6-x)+C