What is the integral of (2x-1)^{-1/2} ?

The integral of (2x-1)^{-1/2} is (2x-1)^{1/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x-1)^{-1/2}

\int (2 x - 1)^{- \frac{1}{2}} d x

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

Solution steps

\int (2 x - 1)^{- \frac{1}{2}} d x

Apply u-substitution

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

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

Apply the Power Rule

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

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

Substitute back

u = 2 x - 1

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

Simplify

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

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

Add a constant to the solution

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

\int k (1 - x) d x

\int 15 sin (x) d x

\int \frac{sin ( z ) cos ( z )}{1 + sin ^{4} ( z )} d z

\int a - x d x

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

What is the integral of (2x-1)^{-1/2} ?
The integral of (2x-1)^{-1/2} is (2x-1)^{1/2}+C