# How do you integrate int (1) / (sqrt(1 + x))?

May 28, 2018

$\int \frac{1}{\sqrt{x + 1}} \mathrm{dx} = 2 \sqrt{x + 1} + c$

#### Explanation:

$\int \frac{1}{\sqrt{x + 1}} \mathrm{dx} = 2 \int \frac{\left(x + 1\right) '}{2 \sqrt{x + 1}} \mathrm{dx} =$

$2 \int \left(\sqrt{x + 1}\right) ' \mathrm{dx} = 2 \sqrt{x + 1} + c$ $\textcolor{w h i t e}{a a}$ , $c$$\in$$\mathbb{R}$

May 28, 2018

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

#### Explanation:

This function is very close to $\sqrt{\setminus \frac{1}{x}}$, whose integral is $2 \sqrt{x}$. In fact,

$\setminus \frac{d}{\mathrm{dx}} 2 \sqrt{x} = 2 \setminus \frac{d}{\mathrm{dx}} \sqrt{x} = 2 \setminus \frac{1}{2 \sqrt{x}} = \setminus \frac{1}{\sqrt{x}}$

In our integral, you can substitute $t = x + 1$, which implies $\mathrm{dt} = \mathrm{dx}$, since this is only a translation. So, you'd have

$\setminus \int \setminus \frac{1}{\sqrt{t}} \mathrm{dt} = 2 \sqrt{t} + C = 2 \sqrt{1 + x} + C$