Solve $\int {\left\{ {\dfrac{{\log \sqrt x }}{{\left( {3x} \right)}}} \right\}} dx = $
$\eqalign{
  & 1)\left( {\dfrac{1}{3}} \right){\left( {\log \sqrt x } \right)^2} + c \cr
  & 2)\left( {\dfrac{2}{3}} \right){\left( {\log \sqrt x } \right)^2} + c \cr
  & 3)\left( {\dfrac{2}{3}} \right){\left( {\log x} \right)^2} + c \cr
  & 4)\left( {\dfrac{1}{3}} \right){\left( {\log x} \right)^2} + c \cr} $

Hint: We need to find the integral of the given expression. The expression contains a logarithmic function and root in the numerator and a variable in the denominator.
First, we can go ahead by simplifying the root. We need to express the root in a simpler way and in a way for which a formula can be applied.
Then, we can take all the constant out. We need to keep only the logarithmic and the algebraic function inside the integral.
Further, we can apply the suitable method and formula to find the value of the integral.
Formulas used to solve the problem are:
$\eqalign{
  & \log {x^n} = n\log x \cr
  & \dfrac{d}{{dx}}\log x = \dfrac{1}{x} \cr
  & \int {xdx = \dfrac{{{x^2}}}{2}} + c \cr} $

Complete step-by-step answer:
The given expression is, $\int {\left\{ {\dfrac{{\log \sqrt x }}{{\left( {3x} \right)}}} \right\}} dx$
Let, $I = \int {\left\{ {\dfrac{{\log \sqrt x }}{{\left( {3x} \right)}}} \right\}} dx$
Where, $\sqrt x $ can be expressed as ${x^{\dfrac{1}{2}}}$
$ \Rightarrow I = \int {\left\{ {\dfrac{{\log {x^{\dfrac{1}{2}}}}}{{\left( {3x} \right)}}} \right\}} dx$
Now, by applying the above formula,
$ \Rightarrow I = \int {\left\{ {\dfrac{{\dfrac{1}{2}\log x}}{{\left( {3x} \right)}}} \right\}} dx$
By taking the constant term out, we get
$ \Rightarrow I = \dfrac{1}{6}\int {\dfrac{{\log x}}{x}} dx$
Now, we shall use the substitution method to solve the above integral.
Let, $\log x = t$
Differentiating on both sides,
$\dfrac{1}{x}dx = dt$
Now, by substituting these values in the integral, we get,
$I = \dfrac{1}{6}\int {t.dt} $
Now, by using the formula,
$I = \dfrac{1}{6}\left( {\dfrac{{{t^2}}}{2}} \right) + c$
Substituting for $t$,
$I = \dfrac{1}{6}\left( {\dfrac{{{{\left( {\log x} \right)}^2}}}{2}} \right) + c$
By rearranging the equation, we get
$\eqalign{
  & \Rightarrow I = \dfrac{1}{3}{\left( {\dfrac{1}{2}\log x} \right)^2} \cr
  & \Rightarrow I = \dfrac{1}{3}{\left( {\log \sqrt x } \right)^2} \cr} $
Therefore, the final answer is $\dfrac{1}{3}{\left( {\log \sqrt x } \right)^2}$
Hence, option (1) gives us the correct answer
So, the correct answer is “Option 1”.

Note: There are two parts to this problem. Simplification and using the substitution method to solve for the integral
Make sure you use the proper formulas, memorize the required integral and derivative formulas.
Logarithmic function when mixed with an algebraic function can be either solved by integration by parts or by the substitution method.