What is the integral of 1/(sqrt(9-2x)) ?

The integral of 1/(sqrt(9-2x)) is -sqrt(9-2x)+C

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integral of 1/(sqrt(9-2x))

\int \frac{1}{9 - 2 x} d x

- 9 - 2 x + C

Solution steps

\int \frac{1}{9 - 2 x} d x

Apply u-substitution

= \int - \frac{1}{2 u} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

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

Substitute back

u = 9 - 2 x

= - \frac{1}{2} \cdot 2 (9 - 2 x)^{\frac{1}{2}}

Simplify

- \frac{1}{2} \cdot 2 (9 - 2 x)^{\frac{1}{2}} : - 9 - 2 x

= - 9 - 2 x

Add a constant to the solution

= - 9 - 2 x + C

What is the integral of 1/(sqrt(9-2x)) ?
The integral of 1/(sqrt(9-2x)) is -sqrt(9-2x)+C