Đáp án:
Đề bài: ${I_8} = \int\limits_{\dfrac{1}{2}}^1 {\dfrac{1}{{x\sqrt {(3 + 4x)(3 - 2x)} }}dx} $
Đặt $t=\dfrac{1}{x} \rightarrow dx=\dfrac{-dt}{t^2}$
\[x=\frac{1}{2} \rightarrow t=2\] \[x=1 \rightarrow t=1\] \[\to I_8=\int_{1}^{2}\dfrac{tdt}{t^2\sqrt{\left(3+\dfrac{4}{t} \right)\left(3-\dfrac{2}{t} \right)}}=\int_{1}^{2}\dfrac{dt}{\sqrt{(3t+4)(3t-2)}}=\int_{1}^{2}\dfrac{dt}{\sqrt{(3t+1)^2-9}}\]
Đặt $u=3t+1+\sqrt{(3t+1)^2-9} \rightarrow \dfrac{du}{3u}=\dfrac{dt}{\sqrt{(3t+1)^2-9}}$
\[t=1 \rightarrow u=4+\sqrt{7}\] \[t=2 \rightarrow u=7+2\sqrt{10}\] \[\to I_8=\frac{1}{3}\int_{4+\sqrt{7}}^{7+2\sqrt{10}}\frac{du}{u}=\frac{1}{3}\ln{u}\Big|_{4+\sqrt{7}}^{7+2\sqrt{10}}=\frac{1}{3}\ln{\left(\frac{7+2\sqrt{10}}{4+\sqrt{7}} \right)}\]
