# How do you integrate sqrt(4x² + 1)?

Aug 14, 2018

$\therefore I = \frac{x}{2} \sqrt{4 {x}^{2} + 1} + \frac{1}{4} \ln | 2 x + \sqrt{4 {x}^{2} + 1} | + c$

#### Explanation:

We know that,

color(red)((1)intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c

Let.

$I = \int \sqrt{4 {x}^{2} + 1} \mathrm{dx}$

Substitute $2 x = u \implies x = \frac{u}{2} \implies \mathrm{dx} = \frac{1}{2} \mathrm{du}$

$I = \int \sqrt{{u}^{2} + 1} \cdot \frac{1}{2} \mathrm{du}$

$\therefore I = \frac{1}{2} \int \sqrt{{u}^{2} + {1}^{2}} \mathrm{du}$

Using $\left(1\right)$ we get

$I = \frac{1}{2} \left\{\frac{u}{2} \sqrt{{u}^{2} + {1}^{2}} + {1}^{2} / 2 \ln | u + \sqrt{{u}^{2} + {1}^{2}} |\right\} + c$

Subst. back $u = 2 x$

$I = \frac{1}{2} \left\{\frac{2 x}{2} \sqrt{4 {x}^{2} + 1} + \frac{1}{2} \ln | 2 x + \sqrt{4 {x}^{2} + 1} |\right\} + c$

$\therefore I = \frac{x}{2} \sqrt{4 {x}^{2} + 1} + \frac{1}{4} \ln | 2 x + \sqrt{4 {x}^{2} + 1} | + c$

Aug 14, 2018

$I = \frac{x}{2} \sqrt{4 {x}^{2} + 1} + \frac{1}{4} \ln | 2 x + \sqrt{4 {x}^{2} + 1} | + C$

#### Explanation:

Here,

$I = \int \sqrt{4 {x}^{2} + 1} \mathrm{dx} \ldots \ldots \ldots \ldots \ldots . \to \left(1\right)$

$I = \int \sqrt{4 {x}^{2} + 1} \cdot 1 \mathrm{dx}$

Using Integration by parts:

$I = \sqrt{4 {x}^{2} + 1} \int 1 \mathrm{dx} - \int \left(\frac{d}{\mathrm{dx}} \left(\sqrt{4 {x}^{2} + 1}\right) \int 1 \mathrm{dx}\right) \mathrm{dx}$

$\therefore I = \sqrt{4 {x}^{2} + 1} \cdot x - \int \frac{8 x}{2 \sqrt{4 {x}^{2} + 1}} \times x \mathrm{dx}$

$\therefore I = x \sqrt{4 {x}^{2} + 1} - \int \frac{4 {x}^{2}}{\sqrt{4 {x}^{2} + 1}} \mathrm{dx}$

$\therefore I = x \sqrt{4 {x}^{2} + 1} - \int \frac{4 {x}^{2} + 1 - 1}{\sqrt{4 {x}^{2} + 1}} \mathrm{dx}$

$\therefore I = x \sqrt{4 {x}^{2} + 1} - \int \sqrt{4 {x}^{2} + 1} \mathrm{dx} + \int \frac{1}{\sqrt{4 {x}^{2} + 1}} \mathrm{dx}$

$\therefore I = x \sqrt{4 {x}^{2} + 1} - I + \int \frac{1}{\sqrt{{\left(2 x\right)}^{2} + 1}} \to \left[\because \left(1\right)\right]$

$\therefore I + I = x \sqrt{4 {x}^{2} + 1} + \frac{1}{2} \ln | 2 x + \sqrt{{\left(2 x\right)}^{2} + 1} | + c$

$\therefore 2 I = x \sqrt{4 {x}^{2} + 1} + \frac{1}{2} \ln | 2 x + \sqrt{4 {x}^{2} + 1} | + c$

$\therefore I = \frac{x}{2} \sqrt{4 {x}^{2} + 1} + \frac{1}{4} \ln | 2 x + \sqrt{4 {x}^{2} + 1} | + C$

Aug 14, 2018

$\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx} = \frac{1}{4} {\sinh}^{- 1} x + \frac{1}{2} x \sqrt{4 {x}^{2} + 1} + C$

#### Explanation:

One method is to use a hyperbolic substitution with $x = \frac{1}{2} \sinh u$

Then:

$\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx} = \int \setminus \sqrt{{\sinh}^{2} u + 1} \setminus \frac{\mathrm{dx}}{\mathrm{du}} \setminus \mathrm{du}$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \int \setminus \sqrt{{\sinh}^{2} u + 1} \frac{1}{2} \cosh u \setminus \mathrm{du}$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \int \setminus \sqrt{{\cosh}^{2} u} \frac{1}{2} \cosh u \setminus \mathrm{du}$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \int \setminus \frac{1}{2} {\cosh}^{2} u \setminus \mathrm{du}$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \int \setminus \frac{1}{4} \left(1 + \cosh 2 u\right) \setminus \mathrm{du}$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \frac{1}{4} u + \frac{1}{8} \sinh 2 u + C$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \frac{1}{4} u + \frac{1}{4} \sinh u \cosh u + C$

$\textcolor{w h i t e}{\int \setminus \sqrt{4 {x}^{2} + 1} \setminus \mathrm{dx}} = \frac{1}{4} {\sinh}^{- 1} x + \frac{1}{2} x \sqrt{4 {x}^{2} + 1} + C$