Ta có: $2x+1+\sqrt {4x + 1} = \frac{1}{2}(4x + 2 + 2\sqrt {4x + 1} ) = \frac{1}{2}{(\sqrt {4x + 1} + 1)^2}$$\Leftrightarrow I= \int\limits_2^6 {\frac{{dx}}{{2x + 1 + \sqrt {4x + 1} }}} = 2\int\limits_2^6 {\frac{{dx}}{{{{(\sqrt {4x + 1} + 1)}^2}}}} $Đặt $t = \sqrt {4x + 1} + 1 \Leftrightarrow {(t - 1)^2} = 4x + 1 \Leftrightarrow \frac{1}{2}(t - 1)dt = dx$\[x = 6 \Rightarrow t = 6\]\[x = 2 \Rightarrow t = 4\]$ \Rightarrow I = \int\limits_4^6 {\frac{{(t - 1)dt}}{{{t^2}}}} = \int\limits_4^6 {(\frac{1}{t}} - \frac{1}{{{t^2}}})dt = \ln \frac{3}{2} - \frac{1}{{12}}$
Ta có: $2x+1+\sqrt {4x + 1} = \frac{1}{2}(4x + 2 + 2\sqrt {4x + 1} ) = \frac{1}{2}{(\sqrt {4x + 1} + 1)^2}$$\Leftrightarrow I= \int\limits_2^6 {\frac{{dx}}{{2x + 1 + \sqrt {4x + 1} }}} = 2\int\limits_2^6 {\frac{{dx}}{{{{(\sqrt {4x + 1} + 1)}^2}}}} $Đặt $t = \sqrt {4x + 1} + 1 \Leftrightarrow {(t - 1)^2} = 4x + 1 \Leftrightarrow \frac{1}{2}(t - 1)dt = dx$\[x = 6 \Rightarrow t = 6\]\[x = 2 \Rightarrow t = 4\]$ \Rightarrow I = \int\limits_3^5 {\frac{{(t - 1)dt}}{{{t^2}}}} = \int\limits_3^5 {(\frac{1}{t}} - \frac{1}{{{t^2}}})dt = \ln \frac{3}{2} - \frac{1}{{12}}$
Ta có: $2x+1+\sqrt {4x + 1} = \frac{1}{2}(4x + 2 + 2\sqrt {4x + 1} ) = \frac{1}{2}{(\sqrt {4x + 1} + 1)^2}$$\Leftrightarrow I= \int\limits_2^6 {\frac{{dx}}{{2x + 1 + \sqrt {4x + 1} }}} = 2\int\limits_2^6 {\frac{{dx}}{{{{(\sqrt {4x + 1} + 1)}^2}}}} $Đặt $t = \sqrt {4x + 1} + 1 \Leftrightarrow {(t - 1)^2} = 4x + 1 \Leftrightarrow \frac{1}{2}(t - 1)dt = dx$\[x = 6 \Rightarrow t = 6\]\[x = 2 \Rightarrow t = 4\]$ \Rightarrow I = \int\limits_
4^
6 {\frac{{(t - 1)dt}}{{{t^2}}}} = \int\limits_
4^
6 {(\frac{1}{t}} - \frac{1}{{{t^2}}})dt = \ln \frac{3}{2} - \frac{1}{{12}}$