Let I = ∫dx/{(a-x)*(b-x)}

      1/{(a-x)*(b-x)} = A/(a-x) + B/(b-x)        {using partial fraction}

=> 1/{(a-x)*(b-x)} = {A(b-x) + B(a-x)}/{(a-x)*(b-x)}

=> 1 = A(b-x) + B(a-x)

=> 1 = Ab - Ax + Ba - Bx

=> 1 = -(A + B)x + (Ab + Ba)

=> A + B = 0 and Ab + Ba = 1

=> A = -B

and Ab + Ba = 1

=> -Bb + Ba = 1

=> B(a - b) = 1

=> B = 1/(a - b)

and A = -1/(a - b)

=> A = 1/(b - a)

Now,

1/{(a-x)*(b-x)} = 1/{(a-x)*(b-a)} + 1/{(b-x)*(a-b)}

So, I = ∫dx/{(a-x)*(b-x)}dx

=> I = ∫[1/{(a-x)*(b-a)} + 1/{(b-x)*(a-b)}]dx

=> I = {1/(b-a)}∫1/(a-x) dx + {1/(a-b)}∫1/(b-x) dx

=> I = -log(a-x)/(b-a) - log(b-x)/(a-b) + C

=> I = log(a-x)/(a-b) + log(b-x)/(b-a) + C

So, ∫dx/{(a-x)*(b-x)}dx = log(a-x)/(a-b) + log(b-x)/(b-a) + C