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$\begin{array}{l}
1)\,\,\,I = \,\,\int\limits_0^{\frac{1}{2}} {\frac{{{x^4}}}{{{x^2} - 1}}dx = } \int\limits_0^{\frac{1}{2}} {\left( {{x^2} + 1 + \frac{1}{{{x^2} - 1}}} \right)} dx\\
\,\,\,\,\,\,\,\,\, = \,\,\,\left[ {\frac{{{x^3}}}{3} + x + \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right|} \right]_0^{\frac{1}{2}} = \frac{{13}}{{24}} - \frac{1}{2}\ln 3
\end{array}$
$2)\,\,\,\,I_t = \int\limits_0^t {\frac{{\tan^4x\left( {1 + \tan^2x} \right)}}{{1 - \tan^2x}}dx\,\,\, = \,\,\int\limits_0^t {\frac{{\tan^4xd\left( {\tan x} \right)}}{{1 - \tan^2 x}}} } $(do $(\tan x)'=\frac{1}{\cos^2x}=1+\tan^2x$)
Đặt $u = \tan x\,\,\,\,\, \Rightarrow \,\,\,\,du = d\left( {\tan x} \right)\,\,,\,\,\left\{ \begin{array}{l}
x = 0\,\,\, \Rightarrow \,\,u = 0\\
x = t\,\,\, \Rightarrow \,\,u = \tan t
\end{array} \right.$
Suy ra : $I = \,\, - \int\limits_0^{\tan t} {\frac{{{u^4}}}{{{u^2} - 1}}} du$
$\begin{array}{l}
= \left[ {\frac{{{u^3}}}{3} + u + \frac{1}{2}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right|} \right]_{\tan t}^{0}\\
= \frac{1}{2}\ln \left| {\tan\left( {t + \frac{\pi }{4}} \right)} \right| - \frac{{\tan ^3t}}{3} - \tan t
\end{array}$
$\begin{array}{l}
\forall t \in \left( {0,\frac{\pi }{4}} \right)\,\,\,:\,\,\,\frac{{\tan^4x}}{{\cos 2x}} > 0\,\,\,\,\, \Rightarrow \,\,\,{I_{(t)}} > 0\\
\Rightarrow \,\,\,\frac{1}{2}\ln \left| {\tan\left( {t + \frac{\pi }{4}} \right)} \right| > \frac{{\tan^3t}}{3} + \tan t\\
\Rightarrow \,\,\ln \left[ {\tan\left( {t + \frac{\pi }{4}} \right)} \right] > \frac{2}{3}\left( {\tan^3t + 3\tan t} \right)\\
\Rightarrow \,\,\tan\left( {t + \frac{\pi }{4}} \right) > {e^{\frac{2}{3}\left( {\tan^3t + 3\tan t} \right)}}
\end{array}$
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