`(2x)/(x^2-9)lt=1/(x+2)`

The solution set of (10x)/(x^(2)+9)lt=1 is

Solve 3-(x)/(4)lt2x-9lt12-(x)/(2),x inZ

If f(x)={:{((x^(2)-9)/(x-3)", for " 0 lt x lt 3),(x+3", for " 3 le x lt 6),((x^(2)-9)/(x+3)", for " 6 le x lt 9):} , then f is

Compute Lt_(x to 3) (x^2 - 9)/(x^3 - 6x^2 + 9x + 1)

y = cos^(-1)((2x)/(1+x^2)), -1 lt x lt 1 .

Integrate : int "tan"^(-1)(2x)/(1-x^(2))dx(-1 lt x lt 1)

The solution set of ((x^(2)+x+1)(x^(2))log(x))/((x+1)(x+9))lt 0 is

The solution set of ((x^(2)+x+1)(x^(2))log(x))/((x+1)(x+9))lt 0 is

For x in(0,1) arrange f_(1)(x) = (1)/(9-x^(2)), f_(2)(x) = (1)/(9-2x^(2)) and f_(3)(x) = (1)/(9-x^(2)-x^(3)) in ascending order and hence prove that 1/6 ln2 lt int_(0)^(1)(1)/(9-x^(2)-x^(3)) dx lt (1)/(6sqrt(2)) ln 5 .

For x in(0,1) arrange f_(1)(x) = (1)/(9-x^(2)), f_(2)(x) = (1)/(9-2x^(2)) and f_(3)(x) = (1)/(9-x^(2)-x^(3)) in ascending order and hence prove that 1/6 ln2 lt int_(0)^(1)(1)/(9-x^(2)-x^(3)) dx lt (1)/(6sqrt(2)) ln 5 .