Question

solve the problem given ​

Answer 1

Explanation:

I Used tan(u+v)=tanu+tanv1−tanu⋅tanv identity for decomposing arctan(2x−11+x−x2).

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

3) After summing 2 integrals, I found result.

Then according to the question:

I=∫10arctan[2x−11+x−x2]⋅dx

=∫10arctan(2x−11−x⋅(x+1))⋅dx

=∫10arctanx⋅dx+∫10arctan(x−1)⋅dx

After using x=1−u an dx=−dutransforms,

I=∫01arctan(1−u)⋅(−du)+∫01arctan(−u)⋅(−du)

=∫01arctan(u−1)⋅du+∫01arctanu⋅du

=−∫10arctanu⋅du-∫10arctan(u−1)⋅du

=−∫10arctanx⋅dx-∫10arctan(x−1)⋅dx

After summing 2 integrals,

2I=0, and value of I=0.

Hope this would be helpful to you.

Also refer to the attachment: