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}$
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}}\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}$
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}$