# How do you evaluate the integral int sqrt(4+x^2)?

##### 1 Answer
May 29, 2018

$I = x \sqrt{1 + {x}^{2} / 4} + 2 {\sinh}^{- 1} \left(\frac{x}{2}\right) + C$

#### Explanation:

$I = \int \sqrt{4 + {x}^{2}} \textcolor{red}{\mathrm{dx}}$

You can use identity:

• ${\cosh}^{2} y - {\sinh}^{2} y = 1 \implies {\cosh}^{2} y = 1 + {\sinh}^{2} y$

So let:

• ${x}^{2} = 4 {\sinh}^{2} y$

• $\implies 2 x \setminus \mathrm{dx} = 8 \sinh y \setminus \cosh y \setminus \mathrm{dy}$

$\implies I = \int \sqrt{4 + 4 {\sinh}^{2} y} \setminus \frac{8 \sinh y \setminus \cosh y \setminus \mathrm{dy}}{2 x}$

$= \int 2 \cosh y \setminus \frac{8 \sinh y \setminus \cosh y \setminus \mathrm{dy}}{2 \cdot 2 \sinh y}$

$= 4 \int {\cosh}^{2} y \setminus \mathrm{dy}$

$= 4 \int \frac{\cosh 2 y + 1}{2} \setminus \mathrm{dy}$

$= \sinh 2 y + 2 y + C q \quad \triangle$

Considering:

• $\textcolor{red}{\sinh 2 y = 2 \sinh y \cosh y}$

• ${x}^{2} = 4 {\sinh}^{2} y \implies \textcolor{red}{x = 2 \sinh y} \textcolor{red}{\implies y = {\sinh}^{- 1} \left(\frac{x}{2}\right)}$

• ${\cosh}^{2} y = 1 + {\sinh}^{2} y \implies \textcolor{red}{\cosh y = \sqrt{1 + {x}^{2} / 4}}$

Then $\triangle$ becomes:

$I = 2 \cdot \frac{x}{2} \cdot \sqrt{1 + {x}^{2} / 4} + 2 {\sinh}^{- 1} \left(\frac{x}{2}\right) + C$

$= x \sqrt{1 + {x}^{2} / 4} + 2 {\sinh}^{- 1} \left(\frac{x}{2}\right) + C$