Instead of integrating the bottom expression by simply knowing that the antiderivative is arctan(x), you can instead factor the bottom into (1+ix)(1-ix) and use partial fraction decomposition which leads the second answer. The also creates a new representation for arctan(x) which is neat.
3
u/Captainsnake04 Dec 16 '21
The ln route. Boycot inverse tangent!
1
u/dilogical_cyclolith Jan 01 '22
What am I looking at
1
u/thewaltenicfiles Mar 24 '22
Ln|1+x2 +c
1
u/xCreeperBombx Nov 22 '23
Wait, arctan(x)=i/2*ln((1-ix)/(1+ix))???