Tính \(I = \int {{{{{\ln }^2}xdx} \over {x\root 3 \of {2 - \ln x} }}} \)
A.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} - 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
B.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
C.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} - {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
D.\(I = {3 \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
A.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} - 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
B.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
C.\(I = {{ - 3} \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} - {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
D.\(I = {3 \over 8}\root 3 \of {{{\left( {2 - \ln x} \right)}^8}} + {{12} \over 5}\root 3 \of {{{\left( {2 - \ln x} \right)}^5}} + 6\root 3 \of {{{\left( {2 - \ln x} \right)}^2}} + C\)
Loga
Hóa Học lớp 12
