Correct Answer - 108
`I=int_(0)^(1)(sin^(-1)sqrt(x))/(x^(2)-x+1)dx`……………1
`I=int_(0)^(1)(sin^(-1)sqrt(1-x))/(x^(2)-x+1)dx=int_(0)^(1)(cos^(-1)sqrt(x))/(x^(2)-x+1)dx`…………..2
on adding 1 and 2 we get
`2I=int_(0)^(1)(sin^(-1)sqrt(x)cos^(-1)sqrt(x))/(x^(2)-x+1) dx`
`=(pi)/2 int_(0)^()(dx)/(x^(2)-x+1)dx`
` =(pi)/2 int_(0)^(1)(dx)((x-1/2)^(2)+((sqrt(3))/2)^(2))dx`
`:.2I=(pi)/2 1/(((sqrt(3))/2))[tan^(-1)((2x-1)/(sqrt(3)))]_(0)^(1)=(pi)^(2)/(3sqrt(3))`
Hence `I=(pi^(2))/(6sqrt(3))=(pi^(2))/(sqrt(108))=(pi^(2))/(sqrt(n))`