$y'=(\dfrac{x^2+x+4}{x+1})'$
$=\dfrac{(x^2+x+4)'(x+1)-(x^2+x+4)(x+1)'}{(x+1)^2}$
$=\dfrac{(2x+1)(x+1)-(x^2+x+4)}{(x+1)^2}$
$=\dfrac{2x^2+3x+1-x^2-x-4}{(x+1)^2}$
$=\dfrac{x^2+2x-3}{(x+1)^2}$
.
+) $y'>0$
<=>$ \dfrac{x^2+2x-3}{(x+1)^2}>0$
Ta có bảng
$\text{ x -∞ -3 -1 1 +∞ }$
$\text{ f(x) + 0 - || - 0 + }$
$=> x ∈ (-∞; -3) ∪ (1; +∞)$