Answer
$$\frac{1}{{12}}\left( {2\ln x + 1} \right)\sqrt {4\ln x - 1} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\ln x}}{{x\sqrt {4\ln x - 1} }}dx} \cr
& {\text{Make an appropiate }}u{\text{ - substitution }} \cr
& u = \ln x,\,\,\,\,\,\,\,du = \frac{1}{x}dx,\,\,\,\,\,\,dx = xdu \cr
& {\text{write in terms of }}u \cr
& \int {\frac{{\ln x}}{{x\sqrt {4\ln x - 1} }}dx} = \int {\frac{u}{{x\sqrt {4u - 1} }}\left( {xdu} \right)} \cr
& = \int {\frac{u}{{\sqrt {4u - 1} }}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 105}} \cr
& \left( {105} \right):\,\,\,\,\,\,\int {\frac{{udu}}{{\sqrt {a + bu} }} = \frac{2}{{3{b^2}}}\left( {bu - 2a} \right)\sqrt {a + bu} + C} \cr
& {\text{take }}a = - 1,\,\,\,b = 4 \cr
& \int {\frac{u}{{\sqrt {4u - 1} }}du} = \frac{2}{{3{{\left( 4 \right)}^2}}}\left( {4u - 2\left( { - 1} \right)} \right)\sqrt {4u - 1} + C \cr
& {\text{simplifying}} \cr
& = \frac{2}{{48}}\left( {4u + 2} \right)\sqrt {4u - 1} + C \cr
& = \frac{1}{{12}}\left( {2u + 1} \right)\sqrt {4u - 1} + C \cr
& {\text{write in terms of }}x{\text{ replace }}\ln x{\text{ for }}u \cr
& = \frac{1}{{12}}\left( {2\ln x + 1} \right)\sqrt {4\ln x - 1} + C \cr} $$
