How do you integrate #x^2/(x^2+1)#?

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Answer 1 · 2016-06-28T07:40:29

#x - arctan x + C#

#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#