Answer
$$\frac{1}{2}{\ln ^2}x{\sin ^{ - 1}}\left( {\ln x} \right) + \frac{{\ln x\sqrt {1 - {{\ln }^2}x} }}{4} - \frac{1}{4}{\sin ^{ - 1}}\left( {\ln x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\left( {\ln x} \right){{\sin }^{ - 1}}\left( {\ln x} \right)}}{x}} dx \cr
& {\text{substitute }}u = \ln x,{\text{ }}du = \frac{1}{x}dx \cr
& \int {\frac{{\left( {\ln x} \right){{\sin }^{ - 1}}\left( {\ln x} \right)}}{x}} dx = \int {u{{\sin }^{ - 1}}udu} \cr
& {\text{A matching integral in a table of integrals at the end of the book is the }} \cr
& {\text{formula }}105 \cr
& \int {{u^n}{{\sin }^{ - 1}}u} du = \frac{1}{{n + 1}}\left( {{u^{n + 1}}{{\sin }^{ - 1}}u - \int {\frac{{{u^{n + 1}}}}{{\sqrt {1 - {u^2}} }}dx} } \right) \cr
& n = 1 \cr
& \int {u{{\sin }^{ - 1}}udu} = \frac{1}{{1 + 1}}\left( {{u^{1 + 1}}{{\sin }^{ - 1}}u - \int {\frac{{{u^{1 + 1}}}}{{\sqrt {1 - {u^2}} }}dx} } \right) \cr
& \int {u{{\sin }^{ - 1}}udu} = \frac{1}{2}\left( {{u^2}{{\sin }^{ - 1}}u - \int {\frac{{{u^2}}}{{\sqrt {1 - {u^2}} }}du} } \right) \cr
& \int {u{{\sin }^{ - 1}}udu} = \frac{1}{2}{u^2}{\sin ^{ - 1}}u - \frac{1}{2}\int {\frac{{{u^2}}}{{\sqrt {1 - {u^2}} }}du} \cr
& {\text{formula 66 }}\int {\frac{{{u^2}}}{{\sqrt {{a^2} - {u^2}} }}dx = } - \frac{u}{2}\sqrt {{a^2} - {u^2}} + \frac{{{a^2}}}{2}{\sin ^{ - 1}}\frac{u}{a} + C,{\text{ }}a = 1 \cr
& \int {u{{\sin }^{ - 1}}udu} = \frac{1}{2}{u^2}{\sin ^{ - 1}}u - \frac{1}{2}\left( { - \frac{u}{2}\sqrt {{a^2} - {u^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{u}{a}} \right) + C \cr
& \int {u{{\sin }^{ - 1}}udu} = \frac{1}{2}{u^2}{\sin ^{ - 1}}u + \frac{u}{4}\sqrt {{a^2} - {u^2}} - \frac{{{a^2}}}{4}{\sin ^{ - 1}}\frac{u}{a} + C \cr
& u = \ln x,{\text{ }}a = 1 \cr
& = \frac{1}{2}{\ln ^2}x{\sin ^{ - 1}}\left( {\ln x} \right) + \frac{{\ln x\sqrt {1 - {{\ln }^2}x} }}{4} - \frac{1}{4}{\sin ^{ - 1}}\left( {\ln x} \right) + C \cr} $$
