Let f(x)= \dfrac {1}{\sqrt{|x|}} when |x|>1 and f(x)=1 when |x| \le 1.Show that f is Lebesgue integrable on \mathbb{R}.

My work:
On |x| \le 1,since f(x)=1,so \int_{[-1,1]} f dm < \infty
but when  |x|>1,I don't think it is integrable as when x>1 , \int_{(1,\infty)} f dm > \int_{[2,n]} f dm
But  \int_{[2,n]} f dm = 2\sqrt{n} -2\sqrt{2},then as n\rightarrow \infty , \int_{[2,n]} f dm \rightarrow \infty and this implies that f in not integrable on (1,\infty).

Am I right?If not,can anyone show me the right way?