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

Hello,

$\dfrac{d^2( \int\limits^x_0 { \dfrac{2}{t^2+3t+1} } \, dt )}{dx^2} = \dfrac{d(\dfrac{2}{x^2+3x+1}) }{dx} \\ = \dfrac{-2(2x+3)}{(x^2+3x+1)^2}$