Cho nguyên hàm \(I = \int {\dfrac{{6{\mathop{\rm tanx}\nolimits} }}{{{{\cos }^2}x\sqrt {3\tan x + 1} }}dx} \) . Giả sử đặt \(u = \sqrt {3\tan x + 1} \) thì ta được:

\(m > 6\)

\(m \ge 6\)

\(m \le 6\)

\(m < 6\)

\({(x + 1)^2} + {(y + 2)^2} + {(z - 3)^2} = 53.\)

\({(x + 1)^2} + {(y + 2)^2} + {(z + 3)^2} = 53.\)

\({(x - 1)^2} + {(y - 2)^2} + {(z - 3)^2} = 53.\)

\({(x - 1)^2} + {(y - 2)^2} + {(z + 3)^2} = 53.\)

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 - \dfrac{1}{2}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {{x^2}\ln \left( {x + 2} \right)} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{2}} \right|_0^1 + \int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

$I = \left. {\dfrac{{{x^2}\ln \left( {x + 2} \right)}}{4}} \right|_0^1 - \dfrac{1}{4}\int\limits_0^1 {\dfrac{{{x^2}}}{{x + 2}}{\rm{d}}x} .$

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b + \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_b^a - \int\limits_a^b {vdu} \)

\(\int\limits_a^b {udv}  = \left. {uv} \right|_a^b - \int\limits_b^a {vdu} \)

\(\left( {1;1; - 1} \right)\)

\(\left( { - 1;1; - 1} \right)\)    

\(\left( {1;1; - \dfrac{1}{3}} \right)\)

\(\left( {3;3; - 1} \right)\)

$I = {x^2}\sin x|_0^\pi  - \int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  + 2\int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  + 2\int\limits_0^\pi  {x\sin xdx} $

$I = {x^2}\sin x|_0^\pi  - 2\int\limits_0^\pi  {x\sin xdx} $