The primitive of 1(x−a)3/2(b−x)1/2 is

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