sir please i want this answer with procedure∫ (1dx)/(a-x)(b-x)

The above integeral can be solved by using Partial Fraction ------------>

 ∫ (1dx)/(a-x)(b-x)  = A/(a -x) + B/(b -x) 
1 = A(b -x) + B(a – x)
Now find the values of A and B.
1. put x=a
1 = A(b – a)
A = 1/(b – a) 
2. put x= b 
B = 1/(a – b)

Hence the integeral,  ∫ (1dx)/(a-x)(b-x) =   ∫1/(b – a) * (a – x)dx  +   ∫ 1/(a – b) *  1/(b – x)dx
                                                             =  1/(a – b)*log(a – x)  – 1/(a – b) * log(b – x)  + c
is the answer.