Cho \[2\sqrt 3 m - \mathop \smallint \limits_0^1 \frac{{4{x^3}}}{{{{\left( {{x^4} + 2} \right)}^2}}}dx = 0\]. Khi đó \[144{m^2} - 1\;\]bằng:

A.\[I = - \mathop \smallint \limits_{\sqrt 2 }^{\frac{2}{{\sqrt 3 }}} \frac{{{t^2}}}{{{t^2} - 1}}dt\]

B. \[I = \mathop \smallint \limits_2^3 \frac{{{t^2}}}{{{t^2} + 1}}dt\]

C. \[I = \mathop \smallint \limits_{\sqrt 2 }^{\frac{2}{{\sqrt 3 }}} \frac{{{t^2}}}{{{t^2} - 1}}dt\]

D. \[I = \mathop \smallint \limits_2^3 \frac{t}{{{t^2} + 1}}dt\]

A.\[I = - \frac{1}{4}{\pi ^4}\]

B. \[I = - {\pi ^4}\]

C. \[I = 0\]

D. \[I = - \frac{1}{4}\]Trả lời:

A.\[I = 2\mathop \smallint \limits_8^9 \sqrt u du\]

B. \[I = \frac{1}{2}\mathop \smallint \limits_8^9 \sqrt u du\]

C. \[I = \mathop \smallint \limits_9^8 \sqrt u du\]

D. \[I = \mathop \smallint \limits_8^9 \sqrt u du\]

A.\[I = \frac{2}{3}\mathop \smallint \limits_1^2 tdt\]

B. \[I = \frac{2}{3}\mathop \smallint \limits_1^2 {t^2}dt\]

C. \[I = \left( {\frac{2}{9}{t^3} + 2} \right)\left| {_1^2} \right.\]

D. \[I = \frac{{14}}{9}\]

A.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 0\]

B.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 1\]

C.\[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = - 1\]

D. \[\mathop \smallint \limits_a^b f'\left( x \right){e^{f\left( x \right)}}dx = 2\]