Giả sử rằng $I = \int\limits_{ - 1}^0 {\dfrac{{3{x^2} + 5x - 1}}{{x - 2}}dx}  = a\ln \dfrac{2}{3} + b$. Khi đó giá trị của $a + 2b$ là :

$m > 6$

$m \ge 6$

$m \le 6$

$m < 6$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\left( { - 1;1; - 1} \right)$    

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

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

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

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

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

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