Question

integrate cot^-1(1-x+x²) dx​

Answer 1

Answer:

cot^-1(1-x+x^2)

=tan^-1(1/(1-x+x^2)

=tan^-1[(x+1-x)/1-x(1-x)

=tan^-1(x)+tan^-1(1-x)

Integration of tan^-1(x)+tan^-1(1-x) dx from 0 to 1 is pi/2-ln2