$T(n) = T(n/4) + T(3n/4) + n$
$a_{1} = 1 , a_{2} = 1$
$b_{1} = 1/4 , b_{2} = 3/4$
$g(n) = n$
$g(u) = u$
$(1/4)^p + (3/4)^p = 1$
when $p = 1, LHS = RHS$
This method gives $T(x) \epsilon \Theta(f(x))$
$f(x) = x^p.( 1 + \int_{1}^{x} g(u)/u^{p+1} )$
$= x.( 1 + \int_{1}^{x} u/u^{2} )$
$= x.( 1 + \int_{1}^{x} 1/u )$
$= x.( 1 + log x )$
$T(n) = \Theta(nlogn)$ , on replacing $x$ witn $n$
1. https://www.youtube.com/watch?v=Gl2v9G0Rn4k