Đặt: $t=1+\dfrac{1}{x} \Rightarrow dt=-\dfrac{1}{x^2}dx$
Đổi cận: $x=\dfrac{1}{2} \Rightarrow t=3$
$x=1 \Rightarrow t=2$
Ta có:
$\int\limits_{\frac{1}{2}}^1\dfrac{1}{x^2}\left(1+\dfrac{1}{x}\right)^{2007}dx$
$=-\int\limits_3^2t^{2007}dt$
$=\dfrac{t^{2008}}{2008}\left|\begin{array}{l}3\\2\end{array}\right.=\dfrac{3^{2008}-2^{2008}}{2008}$