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#

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