What is the integral of 1/(sqrt(bx)) ?

The integral of 1/(sqrt(bx)) is (2sqrt(x))/(sqrt(b))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(bx))

\int \frac{1}{b x} d x

\frac{2 x}{b} + C

Solution steps

\int \frac{1}{b x} d x

\frac{1}{b x} = \frac{1}{b} \frac{1}{x}

= \int \frac{1}{b} \cdot \frac{1}{x} d x

Take the constant out:

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

= \frac{1}{b} \cdot \int \frac{1}{x} d x

Apply the Power Rule

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

= \frac{1}{b} \cdot 2 x^{\frac{1}{2}}

Simplify

\frac{1}{b} \cdot 2 x^{\frac{1}{2}} : \frac{2 x}{b}

= \frac{2 x}{b}

Add a constant to the solution

= \frac{2 x}{b} + C

h \to 0 lim ((1 + h)^{\frac{1}{h}})

n = 1 \sum \infty n^{- \frac{1}{2}}

y = \frac{x ^{4}}{1 - x ^{3}}

x \to 5 + lim (\frac{x - 6}{x - 5})

\int 5 x^{- 6} d x

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