By the fundamental theorem of calculus, the first derivative of
[tex]y=\displaystyle\int_0^x\frac2{1+3t+t^2}\,\mathrm dt[/tex]
is
[tex]y'=\displaystyle\frac{\mathrm d}{\mathrm dx}\int_0^x\frac2{1+3t+t^2}\,\mathrm dt=\frac2{1+3x+x^2}[/tex]
Then the second derivative is
[tex]y''=\dfrac{(1+3x+x^2)(0)-2(3+2x)}{(1+3x+x^2)^2}=-\dfrac{2(2x+3)}{(1+3x+x^2)^2}[/tex]
From here you can proceed with determining the concavity of [tex]y[/tex] by finding the possible inflection points (when [tex]y''=0[/tex] or undefined). Since the denominator is positive for all [tex]x[/tex], you know that [tex]y''[/tex] is defined for all [tex]x[/tex]. So you're left with solving
[tex]y''=-\dfrac{2(2x+3)}{(1+3x+x^2)^2}=0[/tex]
[tex]\implies 2x+3=0[/tex]
[tex]\implies x=-\dfrac32[/tex]
So you have two intervals to consider, [tex]\left(-\infty,-\dfrac32\right)[/tex] and [tex]\left(-\dfrac32,\infty\right)[/tex]. Check the sign of [tex]y''[/tex] in both of these intervals; if [tex]y''>0[/tex] on some interval, then [tex]y[/tex] is concave upward on that interval.
