Question

How do I integrate?

`int_0^1arctan(x/(x+1))/arctan((1+2x-2x^2)/2)dx`

Answer 1

Shift the interval to be symmetric around `0` then combine the integral with its mirror.

Explanation:

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`