Question

Find the interval on which the curve of y equals the integral from 0 to x of 1 divided by the quantity 1 plus t plus t squared, dt is concave up. It looks like this:
$\int\limits^x_0 {\frac{1}{1 + t + t^2} } \, dt$

Answer 1

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

$-1-2x>0\implies1+2x<0\implies2x<-1\implies x<-\dfrac12$