$$\int {{{\log \sqrt x } \over {3x}}dx} $$ is equal to
- A$${1 \over 3}{\left( {\log \sqrt x } \right)^2} + C$$
- B$${2 \over 3}{\left( {\log \sqrt x } \right)^2} + C$$
- C$${2 \over 3}{\left( {\log x} \right)^2} + C$$
- D$${1 \over 3}{\left( {\log x} \right)^2} + C$$
✓ Correct answer: A
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