Write the function as f(x)=(x+1)sqrt{x^2+2x+5} and then use the Product Rule and the Chain Rule to get the first derivative:
f'(x)=sqrt{x^2+2x+5}+(x+1) * 1/2 * (x^2+2x+5)^{-1/2} * (2x+2)
=(x^2+2x+5+x^2+2x+1)/sqrt{x^2+2x+5}=(2x^2+4x+6)/sqrt{x^2+2x+5}
Now use the Quotient Rule and Chain Rule to find the second derivative:
f''(x)
=(sqrt{x^2+2x+5} * (4x+4)-(2x^2+4x+6) * 1/2 * (x^2+2x+5)^{-1/2} * (2x+2))/(x^2+2x+5)
=((x^2+2x+5)(4x+4)-(2x^2+4x+6)(x+1))/((x^2+2x+5)^{3/2})
=(4x^3+8x^2+20x+4x^2+8x+20-2x^3-4x^2-6x-2x^2-4x-6)/((x^2+2x+5)^{3/2})
So
f''(x)=(2x^3+6x^2+18x+14)/((x^2+2x+5)^{3/2})
Since x^2+2x+5>0 for all real numbers x, the sign (positive or negative) of f''(x) is determined by the sign of 2x^3+6x^2+18x+14, which is determined by the sign of x^3+3x^2+9x+7.
The number x=-1 is a real root of x^3+3x^2+9x+7 since -1+3-9+7=0. If you use long-division of polynomials (or synthetic division), you'll find that x^3+3x^2+9x+7=(x+1)(x^2+2x+7). The quadratic x^2+2x+7 is always positive for all real x.
All of this implies that the sign of f''(x) is determined by the sign of x+1. Since x+1>0 when x > -1 and x+1<0 when x < -1, the same is true for f''(x).
This means the graph of f(x) is concave up for x > -1 (on the interval (-1,infty)) and concave down for x < -1 (on the interval (-infty,-1)).
