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

= \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

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

= - \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

\frac{d}{d x} (ln ∣ 4 x ∣)

\frac{\partial}{\partial y} (sin (x) + e^{x y})

\frac{d}{d x} ((ln (4 x))^{2})

49 y^{^{''}} + 112 y^{^{'}} + 64 y = 0

\int (4 x^{2} - 2 x - 2) d x

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