Question

Find the value of `int_0^1tan^-1((2x-1)/(1+x-x^2))dx`?

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

Answer 1

See below.

Explanation:

) I used `tan(u+v)=(tanu+tanv)/(1-tanu*tanv)` identity for decomposing `arctan((2x-1)/(1+x-x^2))`.

2) I used `x=1-u` transform in `I` integral.

3) After summing 2 integrals, I found result.

`I`=`int_0^1 arctan[(2x-1)/(1+x-x^2)]*dx`

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

=`int_0^1 arctanx*dx`+`int_0^1 arctan(x-1)*dx`

After using `x=1-u` an `dx=-du` transforms,

`I`=`int_1^0 arctan(1-u)*(-du)`+`int_1^0 arctan(-u)*(-du)`

=`int_1^0 arctan(u-1)*du`+`int_1^0 arctanu*du`

=`-int_0^1 arctanu*du`-`int_0^1 arctan(u-1)*du`

=`-int_0^1 arctanx*dx`-`int_0^1 arctan(x-1)*dx`

After summing 2 integrals,

`2I=0`, hence `I=0`.