Integrate (3x^2+x-5) / (x^2-4x)

Thank you!

Possible intermediate steps:
integral (-5+x+3 x^2)/(-4 x+x^2) dx
For the integrand (3 x^2+x-5)/(x^2-4 x), cancel common terms in the numerator and denominator:
= integral (3 x^2+x-5)/((x-4) x) dx
For the integrand (3 x^2+x-5)/((x-4) x), do long division:
= integral (5/(4 x)+47/(4 (x-4))+3) dx
Integrate the sum term by term and factor out constants:
= integral 3 dx+47/4 integral 1/(x-4) dx+5/4 integral 1/x dx
For the integrand 1/(x-4), substitute u = x-4 and du = dx:
= 47/4 integral 1/u du+ integral 3 dx+5/4 integral 1/x dx
The integral of 1/u is log(u):
= (47 log(u))/4+ integral 3 dx+5/4 integral 1/x dx
The integral of 1/x is log(x):
= (47 log(u))/4+(5 log(x))/4+ integral 3 dx
The integral of 3 is 3 x:
= (47 log(u))/4+3 x+(5 log(x))/4+constant
Substitute back for u = x-4:
= 3 x+47/4 log(x-4)+(5 log(x))/4+constant
Which is equivalent for restricted x values to:
= 3 x+47/4 log(4-x)+(5 log(x))/4+constant