If
[tex]y=\displaystyle\int_0^x\frac{\mathrm dt}{1+t+t^2}[/tex]
then by the fundamental theorem of calculus, the derivative is
[tex]y'=\dfrac1{1+x+x^2}[/tex]
and the second derivative is
[tex]y''=-\dfrac{1+2x}{(1+x+x^2)^2}[/tex]
The curve is concave up whenever [tex]y''>0[/tex]. Since the denominator is always positive, you only need to worry about the numerator:
[tex]-\dfrac{1+2x}{(1+x+x^2)^2}>0\implies-1-2x>0[/tex]
Solving for [tex]x[/tex] yields
[tex]-1-2x>0\implies1+2x<0\implies2x<-1\implies x<-\dfrac12[/tex]
