The primitive of 1(x−a)3/2(b−x)1/2 is
1b−ab−xx−a1/2+C
34(b−a)b−xx−a1/2+C
1b−ax−ab−a1/2+C
none of these
putting x−a=(1/t) we have dx=−1/t2dt
and the given integral can be written as
∫−1t2dt(1/t)3/2b−a−1t1/2=−∫dt(b−a)t−1
=−2(b−a)t−1b−a+C=−2b−ab−a−(x−a)x−a+C=2a−bb−xx−a+C