Anyway why divide first derivative of f(x) is due to reverse chain rule.

Can I expand 4(x-1) first then integrate 3/(4x-4)?
∫3/4(x-1) dx
= 3 ∫(4x-4)^-1
= 3.1/4 ln|4x-4|+ c
= 3/4 ln|4x-4|+ c

Integration of 1/f(x) dx in a more general term is ∫f'(x)/f(x)dx =ln lf(x)l +c . So ∫3/4(x-1) dx = 3/4 ∫4/(4x-4)dx to fit into the form of ∫f'(x)/f(x)dx is also equals to 3/4 ∫1/(x-1)dx to fit into ∫f'(x)/f(x)dx. So 3/(4x-4) is not correct as by expanding, you are changing the power base. You have to ensure in ax+b, it cannot be simplified further as I understand your question as if it is asking for 3/4*1/x-1.

∫1/(ax+b) dx =1/a(ln(ax+b)) + c
Therefore if you expand, it is 3/(ax+a) dx where a=4. It doesn't fit into the rule anymore. 3/a* (Ax+b) fits into ∫1/(ax+b) dx. Where A is the coefficient of x which is 1 in this case, and b is -1.

No wonder. Thanks for the explanation!