The numerator is
#2x^2+4x+12=2(x^2+2x+6)#
We perform a long division
#color(white)(aaaa)##x^2+2x+6##color(white)(aaaaa)##|##x^2+7x+10#
#color(white)(aaaa)##x^2+7x+10##color(white)(aaaa)##|##1#
#color(white)(aaaaaa)##0-5x-4#
Therefore
#(2x^2+4x+12)/(x^2+7x+10)=2(1-(5x+4)/(x^2+7x+10))#
We factorise the denominator
#x^2+7x+10=(x+5)(x+2)#
We can perform the decomposition into partial fractions
#(5x+4)/(x^2+7x+10)=(5x+4)/((x+5)(x+2))#
#=A/(x+5)+B/(x+2)=(A(x+2)+B(x+5))/((x+5)(x+2))#
The denominators are the same, we compare the numerators
#5x+4=A(x+2)+B(x+5)#
Let #x=-5#, #=>#, #-21=-3A#, #=>#, #A=7#
Let #x=-2#, #=>#, #-6=3B#, #=>#, #B=-2#
So,
#(2x^2+4x+12)/(x^2+7x+10)=2(1-(7/(x+5)-2/(x+2)))#
Therefore,
#int((2x^2+4x+12)dx)/(x^2+7x+10)=2int(1-(7/(x+5)-2/(x+2)))dx#
#=2intdx-14intdx/(x+5)+4intdx/(x+2)#
#=2x-14ln(|x+5|)+4ln(|x+2|)+C#
