∫sqr(a^2+x^2)dx

设x=it,则
∫sqrt(a^2+x^2)dx
=i∫sqrt(a^2-t^2)dt
=i((1/2)tsqrt(a^2-t^2)+(a^2/2)arcsin(t/a)+C)
=(1/2)itsqrt(a^2-t^2)+i(a^2/2)arcsin(t/a)+C
=(1/2)xsqrt(a^2+x^2)+i(a^2/2)arcsin(-ix/a)+C
=(1/2)xsqrt(a^2+x^2)+(a^2/2)ln(-y+sqrt((x/a)^2+1))