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