Find the interval on which the curve of y equals the integral from 0 to x of 6 divided by the quantity 1 plus 2 times t plus t squared, dt is concave up

Question

Find the interval on which the curve of  is concave up

Need the interval

Answer 1

The definite integral evaluates to y = 6x / (x+1)

dy/dx = ((x+1)(6) - 6x(1) ) / (x+1)² = 6 / (x+1)²

d²y / dx² = -6(2)(x+1) / (x+1)⁴ = -12 / (x+1)³

The interval on which y is concave up corresponds to the interval on which d²y / dx² > 0. Due to the -12 in the numerator, this happens wherever (x+1)³ < 0:

(x+1)³ < 0 on the interval x ∈ (-∞, -1)