∫(0,√3)[ 1/(9+x^2)]dx
let
x= 3tana
dx= 3(seca)^2 da
x=0,a=0
x=√3,a=π/6
∫(0,√3)[ 1/(9+x^2)]dx
=(1/3)∫(0,π/6) da
= (π/18)