Question

∫[(x-a)/(x-b)]^(1/2)*dx

Answer 1

a=b时原积分=x
a≠b时，令t=[(x-a)/(x-b)]^(1/2),则x=(b-a)/(t^2-1)+1,
dx=d[(b-a)/(t^2-1)+1]=2(b-a)t/(t^2-1)^2dt
原积分=2(b-a)*∫[t^2/(t^2-1)^2]dt
令p=∫[t^2/(t^2-1)^2]dt=∫1/(t-1/t)^2dt
q=∫[1/(t^2-1)^2]dt=∫(1/t^2)/(t-1/t)^2dt
则p-q=∫1/(t^2-1)dt=0.5*ln(t^2-1)
p+q=∫(1/t^2+1)/(t-1/t)^2dt
=∫1/(t-1/t)^2d(t-1/t)
=t/(1-t^2)
故p=0.5*[0.5*ln(t^2-1)+t/(1-t^2)]
原积分=(b-a)*[0.5*ln(t^2-1)+t/(1-t^2)]