∫[x=0,∞] (1/(1 + x^2)^2)dx
=(1/2i)(-1/(i+x)+1/(-i+x))dx
=(1/2i)(-log(i+x)+log(x-i))+c
=(i/2)log(x-i)/(x+i)+c
=(i/2)(log1-log-1)=i/2(-2logi)
=(-i)log(e^iπ/2)
=π/2