设√((a-x)/(x-b))=t,则x-b=(a-b)/(t²+1),dx=2(b-a)tdt/(t²+1)²
故 原式=∫dx/[(x-b)√((a-x)/(x-b))]
=∫[2(b-a)tdt/(t²+1)²]/[((a-b)/(t²+1))t]
=(-2)∫dt/(t²+1)
=(-2)arctant+C(C是积分常数)
=(-2)arctan(√((a-x)/(x-b)))+C