# How do you find the indefinite integral of int 1/sqrt(x+1)?

Jan 12, 2017

Let $u = x + 1$. Then $\mathrm{du} = \mathrm{dx}$.

$= \int \frac{1}{\sqrt{u}} \mathrm{du}$

$= \int {u}^{- \frac{1}{2}} \mathrm{du}$

$= 2 {u}^{\frac{1}{2}} + C$

$= 2 \sqrt{x + 1} + C$

Hopefully this helps!