Tính tích phân $I= \int_2^6 \frac{1}{2x+1+\sqrt{4x+1}} dx$

Tính tích phân $I=\displaystyle \int_2^6 \dfrac{1}{2x+1+\sqrt{4x+1}} dx$


Đặt $t=\sqrt{4x+1} \Rightarrow x=\dfrac{t^2-1}{4} \Rightarrow dx= \dfrac{t}{2}dt$

$x=2 \Rightarrow t=3 , \,\,  x=6 \Rightarrow t=5$

$I=\displaystyle \int_3^5 \dfrac{t}{t^2+2t+1} dt=\displaystyle \dfrac{1}{2} \int_3^5 \dfrac{2t+2}{t^2+2t+1} dt-\displaystyle \int_3^5 \dfrac{1}{(t+1)^2} dt$
$=\dfrac{1}{2} \ln |t^2+2t+1| \bigg|_{3}^{5}+\dfrac{1}{t+1}\bigg|_{3}^{5}=\ln \dfrac{3}{2}-\dfrac{1}{12}$