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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 = \int\limits_0^t {\frac{{t{g^4}x\left( {1 + t{g^2}x} \right)}}{{1 - t{g^2}x}}dx\,\,\,
= \,\,\int\limits_0^t {\frac{{t{g^4}xd\left( {tgx} \right)}}{{1 - t{g^2}x}}} } $
Đặt $u = tgx\,\,\,\,\, \Rightarrow \,\,\,\,du = d\left( {tgx} \right)\,\,,\,\,\left\{
\begin{array}{l}
x = 0\,\,\, \Rightarrow \,\,u = 0\\
x = t\,\,\, \Rightarrow \,\,u = tgt
\end{array} \right.$
Suy ra: $I = \,\, - \int\limits_0^{tgt} {\frac{{{u^4}}}{{{u^2} - 1}}} du$
$\begin{array}{l}
= \left[ {\frac{{{u^3}}}{3} + u + \frac{1}{2}\ln \left| {\frac{{u - 1}}{{u + 1}}} \right|}
\right]_0^{tgt}\\
= \frac{1}{2}\ln \left| {tg\left( {t + \frac{\pi }{4}} \right)} \right| - \frac{{t{g^3}t}}{3} - tgt
\end{array}$
$\begin{array}{l}
\forall t \in \left( {0,\frac{\pi }{4}} \right)\,\,\,:\,\,\,\frac{{t{g^4}x}}{{co\,s\,2x}} > 0\,\,\,\,\,
\Rightarrow \,\,\,{I_{(t)}} > 0\\
\Rightarrow \,\,\,\frac{1}{2}\ln \left| {tg\left( {t + \frac{\pi }{4}} \right)} \right| >
\frac{{t{g^3}t}}{3} - tgt\\
\Rightarrow \,\,\ln \left[ {tg\left( {t + \frac{\pi }{4}} \right)} \right] > \frac{2}{3}\left( {t{g^3}t +
3tgt} \right)\\
\Rightarrow \,\,tg\left( {t + \frac{\pi }{4}} \right) > {e^{\frac{2}{3}\left( {t{g^3}t + 3tgt}
\right)}}
\end{array}$
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Đăng bài 09-05-12 09:19 AM
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