# How do you integrate (sqrt(12+4x^2)dx?

Feb 24, 2015

The answer is: $x \sqrt{{x}^{2} + 3} + 3 a r c \sinh \left(\frac{x}{\sqrt{3}}\right) + c$.

The integral can be written:

$\int \sqrt{4 \left(3 + {x}^{2}\right)} \mathrm{dx} = 2 \int \sqrt{3 + {x}^{2}} \mathrm{dx} = \left(1\right)$

Since:

$x = \sqrt{3} \sinh t \Rightarrow \mathrm{dx} = \sqrt{3} \cosh t \mathrm{dt}$, than:

$\left(1\right) = 2 \int \sqrt{3 + 3 {\sinh}^{2} t} \cdot \sqrt{3} \cosh t \mathrm{dt} =$

$= 2 \int \sqrt{3} \sqrt{1 + {\sinh}^{2} t} \sqrt{3} \cosh t \mathrm{dt} = 6 \int \cosh t \cdot \cosh t \mathrm{dt} =$

$= 6 \int {\cosh}^{2} t \mathrm{dt} = \left(2\right)$.

Now remembering that:

$\cosh \left(\frac{\alpha}{2}\right) = \sqrt{\frac{\cosh \alpha + 1}{2}}$,

$\left(2\right) = 6 \int \frac{\cosh 2 t + 1}{2} \mathrm{dt} = \frac{6}{2} \left(\frac{1}{2} \int 2 \cosh 2 t + \int \mathrm{dt}\right) =$

$= 3 \left(\frac{1}{2} \sinh 2 t + t\right) + c = \left(3\right)$

And now, remembering that:

$\sinh 2 \alpha = 2 \sinh \alpha \cosh \alpha$ and
$\cosh \alpha = \sqrt{{\sinh}^{2} \alpha + 1}$,

than:

$\left(3\right) = 3 \left(\frac{1}{2} 2 \sinh t \cosh t + t\right) =$

$= 3 \left(\frac{x}{\sqrt{3}} \sqrt{{x}^{2} / 3 + 1} + a r c \sinh \left(\frac{x}{\sqrt{3}}\right)\right) + c =$

$= 3 \left(\frac{x}{\sqrt{3}} \frac{\sqrt{{x}^{2} + 3}}{\sqrt{3}} + a r c \sinh \left(\frac{x}{\sqrt{3}}\right)\right) + c =$

$= x \sqrt{{x}^{2} + 3} + 3 a r c \sinh \left(\frac{x}{\sqrt{3}}\right) + c$.