How do I integrate?

Let

#I=int_0^1tan^(-1)(x/(x+1))/tan^(-1)((1+2x-2x^2)/2)dx#

Complete the square in the denominator:

#I=int_0^1tan^(-1)(x/(x+1))/tan^(-1)((3-(2x-1)^2)/4)dx#

Apply the substitution #2x-1=u#:

#I=1/2int_-1^1tan^(-1)((1+u)/(3+u))/tan^(-1)((3-u^2)/4)du#

Apply the substitution #v=-u#:

#I=1/2int_-1^1tan^(-1)((1-v)/(3-v))/tan^(-1)((3-v^2)/4)dv#

Add these two together:

#2I=1/2int_-1^1tan^(-1)((1+u)/(3+u))/tan^(-1)((3-u^2)/4)du+1/2int_-1^1tan^(-1)((1-v)/(3-v))/tan^(-1)((3-v^2)/4)dv#

Since the #u# and #v# variables do not matter beyond their respective integrals, lets just rename both #t#:

#4I=int_-1^1tan^(-1)((1+t)/(3+t))/tan^(-1)((3-t^2)/4)dt+int_-1^1tan^(-1)((1-t)/(3-t))/tan^(-1)((3-t^2)/4)dt#

The distributive property of integration allows:

#4I=int_-1^1(tan^(-1)((1+t)/(3+t))/tan^(-1)((3-t^2)/4)+tan^(-1)((1-t)/(3-t))/tan^(-1)((3-t^2)/4))dt#

The denominators are identical:

#4I=int_-1^1(tan^(-1)((1+t)/(3+t))+tan^(-1)((1-t)/(3-t)))/tan^(-1)((3-t^2)/4)dt#

Apply the inverse trigonometric identity #tan^(-1)a+tan^(-1)b=tan^(-1)((a+b)/(1-ab))#:

#4I=int_-1^1tan^(-1)((3-t^2)/4)/tan^(-1)((3-t^2)/4)dt#

Hence:

#4I=int_-1^1dt#

The remaining integral is trivial:

#I=1/2#