I would be inclined to prove it is not rational (and so cannot be an integer) using the "rational root theorem". The rational root theorem say that "if $\frac{m}{n}$ is a root of the polynomial equation $a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= 0$, with all coefficients integers, then the denominator, n, must divide the leading coefficient, $a_n$, and the numerator, m, must divide the constant term, $a_0$".
If $x= \frac{1+ \sqrt{3}}{2}$ then it satisfies the equation $\left(x-\frac{1+ \sqrt{3}}{2}\right)\left(x+ \frac{1- \sqrt{3}}{2}\right)=$$ \left(-\frac{1}{2}- \frac{\sqrt{3}}{2}\right)\left(x-\frac{1}{2}+ \frac{\sqrt{3}}{2}\right)$$= \left(x- \frac{1}{2}\right)^2- \frac{3}{4}= x^2- x- \frac{1}{2}= 0$. Multiplying by 2, that is equivalent to $2x^2- 2x- 1= 0$.
So any rational roots of that must be of the form $\frac{m}{n}$ where m divides -1 (so can only be 1 or -1) and n divides 2 (so can only be 1, -1, 2, and -2). That is the only possible rational roots of that equation are 1, -1, 1/2, or -1/2. The only integer roots are 1 and -1. Since $\frac{1+ \sqrt{3}}{2}$ is not equal to either 1 or -1, it is not an integer.