First we need to evaluate the domain and the first and second derivatives:

#D_h=RR#

#h'(x)=(2x)/(x^2+1)^2#

#h''(x)=(-2(3x^2-1))/(x^2+1)^3=(-6(x+sqrt(3)/3)(x-sqrt(3)/3))/(x^2+1)^3#

Now, #h# is increasing when #h'>0# and decreasing when #h'<0#. Notice that #forall_(x in RR)\ x^2+1>0# so as long and we're interested only in the sing of the derivative we can ommit the denominator.

#2x>0 iff x>0# - function #h# increases when #x in (0;+oo)#

#2x<0 iff x<0# - function #h# decreases when #x in (-oo;0)#

Function #h# is concave up when #h''>0# and concave down when #h''<0#.

*concave up:* #-6(x+sqrt(3)/3)(x-sqrt(3)/3)>0 iff x in (-sqrt(3)/3;sqrt(3)/3)#

*concave down:* #-6(x+sqrt(3)/3)(x-sqrt(3)/3)<0 iff #

#iff x in (-oo;-sqrt(3)/3) cup (sqrt(3)/3;+oo)#