Question

Find the interval on which the curve of y equals the integral from 0 to x of 2 divided by the quantity 1 plus 3 times t plus t squared, dt is concave up.

Answer 1

$y=\displaystyle\int_0^x\frac2{1+3t+t^2}\,\mathrm dt$

By the fundamental theorem of calculus,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac2{1+3x+x^2}$

then by standard differentiation,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=-\dfrac{6+4x}{(1+3x+x^2)^2}$

$y$ is concave upward whenever $\dfrac{\mathrm d^2y}{\mathrm dt^2}>0$. The denominator is always positive, so the sign is determined by the numerator alone:

$6+4x>0\implies x>-\dfrac32$