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

1 Answer
Sep 22, 2017

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.