How do you integrate #x^2/(x^2+1)#?
Risposta:
#x - arctan x + C#
Spiegazione:
#x^2/(x^2+1) = (x^2+1 - 1)/(x^2+1) = 1 - ( 1)/(x^2+1)#
#int 1 - ( 1)/(x^2+1) dx #
#= x - color(red)(int ( 1)/(x^2+1) dx )#
in terms of the red bit, use sub #x = tan t, dx = sec^2 t dt#
this makes it
#int ( 1)/(tan^2 t+1) sec^2 t dt#
# = int ( 1)/(sec^2 t) sec^2 t dt#
#= int dt#
#= arctan x - C#
So the full integral is
#x - arctan x + C#