What is the integral of 1/(sqrt(3-2x)) ?

The integral of 1/(sqrt(3-2x)) is -sqrt(3-2x)+C

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integral of 1/(sqrt(3-2x))

\int \frac{1}{3 - 2 x} d x

- 3 - 2 x + C

Solution steps

\int \frac{1}{3 - 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 = 3 - 2 x

= - \frac{1}{2} \cdot 2 (3 - 2 x)^{\frac{1}{2}}

Simplify

- \frac{1}{2} \cdot 2 (3 - 2 x)^{\frac{1}{2}} : - 3 - 2 x

= - 3 - 2 x

Add a constant to the solution

= - 3 - 2 x + C

\frac{d}{d x} (- \frac{14}{x ^{3}})

(11 + x^{2}) y^{^{'}} - 2 x y = 0

\int \frac{1}{8 x ^{2} - 16 x + 64} d x

\frac{\partial}{\partial t} (s e^{t})

\frac{\partial}{\partial t} (s sin (t))

What is the integral of 1/(sqrt(3-2x)) ?
The integral of 1/(sqrt(3-2x)) is -sqrt(3-2x)+C