What is the antiderivative of # (x^2-1) / (x^2+1)#?

1 Answer
Roy E.
Dec 20, 2016

#x-2 tan^-1 x+C#

Explanation:

#int(x^2-1)/(x^2+1)dx#
#=int(x^2+1-2)/(x^2+1)dx#
#=int1-2/(1+x^2)dx#
#=x-2 tan^-1x+C#.

(If the quoting standard integral #int1/(1+x^2)dx=tan^-1x# is not acceptable, substitute #x=tan u#, #dx=sec^2 u du# and use #sec^2u=1+tan^2u# to reduce the integral to #int 1du#.)

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