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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

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

Solution steps

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

Apply u-substitution

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

Apply the Power Rule

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

Substitute back

u = x + 1

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

Add a constant to the solution

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

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