2(x²+2x+7)/((x-5)(x-1)). x²+2x+7 doesn’t factorise (it has complex zeroes).
There is an x² term in the numerator and denominator so we can reduce the original fraction to:
2+2(x²+2x+7-(x²-6x+5))/((x-5)(x-1))=
2+2(8x+2)/((x-5)(x-1))=2+4(4x+1)/((x-5)(x-1)).
Now we can convert the simpler fraction into partial fractions:
(4x+1)/((x-5)(x-1))=A/(x-1)+B/(x-5) where A and B are constants.
So, A(x-5)+B(x-1)=4x+1.
A+B=4, -5A-B=1, -5A-(4-A)=1, -4A-4=1, -4A=5, A=-5/4 and B=4-A=4+5/4=21/4.
The whole thing becomes:
2+4(-5/(4(x-1))+21/(4(x-5)))=
2-5/(x-1)+21/(x-5).
CHECK
2+(5(x-5)+21(x-1))/((x-5)(x-1))=
2+(-5x+25+21x-21)/((x-5)(x-1))=
2+(16x+4)/((x-5)(x-1))=
(2x²-12x+10+16x+4)/((x-5)(x-1))=
(2x²+4x+14)/(x²-6x+5).