\(B=\int\limits^{2\sqrt{3}}_{\sqrt{5}}\dfrac{xdx}{x^2\sqrt{x^2+4}}\)
Đặt \(\sqrt{x^2+4}=t\Rightarrow x^2=t^2-4\Rightarrow xdx=tdt\) ; \(\left\{{}\begin{matrix}x=\sqrt{5}\Rightarrow t=3\\x=2\sqrt{3}\Rightarrow t=4\end{matrix}\right.\)
\(B=\int\limits^4_3\dfrac{tdt}{\left(t^2-4\right)t}=\int\limits^4_3\dfrac{dt}{\left(t-2\right)\left(t+2\right)}=\dfrac{1}{4}\int\limits^4_3\left(\dfrac{1}{t-2}-\dfrac{1}{t+2}\right)dt\)
\(=\dfrac{1}{4}.\ln\left|\dfrac{t-2}{t+2}\right||^4_3=\dfrac{1}{4}ln\left(\dfrac{5}{3}\right)\)
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