What is the integral of sqrt(x-1/4) ?

The integral of sqrt(x-1/4) is 1/12 (4x-1)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of sqrt(x-1/4)

\int x - \frac{1}{4} d x

\frac{1}{12} (4 x - 1)^{\frac{3}{2}} + C

Solution steps

\int x - \frac{1}{4} d x

Simplify

x - \frac{1}{4} : \frac{1}{2} 4 x - 1

= \int \frac{1}{2} 4 x - 1 d x

Take the constant out:

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

= \frac{1}{2} \cdot \int 4 x - 1 d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x - 1

= \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x - 1)^{\frac{3}{2}}

Simplify

\frac{1}{2} \cdot \frac{1}{4} \cdot \frac{2}{3} (4 x - 1)^{\frac{3}{2}} : \frac{1}{12} (4 x - 1)^{\frac{3}{2}}

= \frac{1}{12} (4 x - 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (4 x - 1)^{\frac{3}{2}} + C

What is the integral of sqrt(x-1/4) ?
The integral of sqrt(x-1/4) is 1/12 (4x-1)^{3/2}+C