$x{\cos ^{ - 1}}x$

Let $I = \int {x{{\cos }^{ - 1}}x} dx = \int {{{\cos }^{ - 1}}x \cdot x} dx$

$= {\cos ^{ - 1}}x \cdot \int x dx - \int {\left( {\cfrac{d}{{dx}}\left( {{{\cos }^{ - 1}}x} \right)\int x dx} \right)} dx$

$= {\cos ^{ - 1}}x\left( {\cfrac{{{x^2}}}{2}} \right) - \int {\cfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}\left( {\cfrac{{{x^2}}}{2}} \right)dx}$

$= \cfrac{{{x^2}}}{2}{\cos ^{ - 1}}x + \cfrac{1}{2}\int {\cfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}dx = \cfrac{{{x^2}}}{2}{{\cos }^{ - 1}}x + \cfrac{1}{2}{I_1}}$

$\Rightarrow$ $I = \cfrac{{{x^2}}}{2}{\cos ^{ - 1}}x + \cfrac{1}{2}{I_1}$

….(i)
Where ${I_1} = \int {\cfrac{{{x^2}}}{{\sqrt {1 - {x^2}} }}dx}$

Put $x = \cos \theta$ $\Rightarrow$ $dx = - \sin \theta d\theta$

$\therefore$ ${I_1} = \int {\cfrac{{{{\cos }^2}\theta \left( { - \sin \theta } \right)}}{{\sqrt {1 - {{\cos }^2}\theta } }}d\theta }$

$= - \int {{{\cos }^2}\theta d\theta } = - \cfrac{1}{2}\int {\left( {1 + \cos 2\theta } \right)d\theta }$

$= - \cfrac{1}{2}\left( {\theta + \cfrac{{\sin 2\theta }}{2}} \right) + {C_1} = - \cfrac{1}{2}\left( {\theta + \cfrac{1}{2} \times 2\sin \theta \cos \theta } \right) + {C_1}$

$= - \cfrac{1}{2}\left( {\theta + \cos \theta \sqrt {\left( {1 - {{\cos }^2}\theta } \right)} } \right) + {C_1}$

$= - \cfrac{1}{2}\left( {{{\cos }^{ - 1}}x + x\sqrt {1 - {x^2}} } \right) + {C_1}$

….(ii)
From (i) and (ii),

we get $I = \left( {2{x^2} - 1} \right)\cfrac{{{{\cos }^{ - 1}}x}}{4} - \cfrac{x}{4}\sqrt {1 - {x^2}} + C$

$\left[ {C = \cfrac{{{C_1}}}{2}} \right]$