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I = $\int\limits_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}$$\frac{dx}{x\sqrt{x^{2}-1}}$
Ta có: $\int\limits_{\frac{2}{\sqrt{3}}}^{\sqrt{2}}$$\frac{xdx}{x^{2}\sqrt{x^{2}-1}}$
Đặt $\sqrt{x^{2}-1}$ = $t$ $=>$ $x^{2}$ $-$ $1$ = $t^{2}$ $=>$ $x^{2}$ = $t^{2}$ $+$ $1$
$=>$ Có $d$($\sqrt{x^{2}-1}$) = $dt$
$<=>$ $\frac{xdx}{\sqrt{x^{2}-1}}$ = $dt$
$<=>$ $xdx$ = $\sqrt{x^{2}-1}$$dt$ = $tdt$
* Đổi cận: $x$ $\frac{2}{\sqrt{3}}$ $\sqrt{2}$
$t$ $\frac{\sqrt{3}}{3}$ $1$
* $\int\limits_{\frac{\sqrt{3}}{3}}^{1}$ $\frac{tdt}{(t^{2}+1)t}$ = $\int\limits_{\frac{\sqrt{3}}{3}}^{1}$ $\frac{dt}{t^{2}+1}$
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