Let, #I=int(arcsinx)^2dx#.
Subst. #arcsinx=t. :. x=sint. :. dx=costdt#.
Hence, #I=intt^2costdt#.
We use the following Rule of Integration by Parts (IBP) :
#intuv'dt=uv-intu'vdt," where, "u=t^2, v'=cost#, so that,
# u'=2t, v=intv'dt=intcostdt=sint#.
#:. I=t^2sint-int2t*sintdt=t^2sint-2I_1#, where,
#I_1=inttsintdt#.
We again use IBP with #u=t, v'=sint; :. u'=1, v=-cost#,
#:. I_1=t(-cost)-int{(1)(-cost)}dt#,
#=-tcost+intcostdt=-tcost+sint#.
Utilising #I_1# in #I#, we get,
#I=t^2sint-2{-tcost+sint}#,
#=t^2sint+2tcost-2sint#,
#=t^2sint+2tsqrt(1-sin^2t)-2sint#.
Reverting from #t" to "x," by, "t=arcsinx, &, sint=x#,
#I=x(arcsinx)^2+2(sqrt(1-x^2))arcsinx-2x+C#.
Feel the Joy of Maths.!