So, we first write this:
#(6x^2+13x+6)/((x+2)(x+1)^2)=A/(x+2)+B/(x+1)+C/(x+1)^2#
By addition we get:
#(6x^2+13x+6)/((x+2)(x+1)^2)=A/(x+2)+(B(x+1)+C)/(x+1)^2=(A(x+1)^2+(x+2)(B(x+1)+C))/((x+2)(x+1)^2)#
#6x^2+13x+6=A(x+1)^2+(x+2)(B(x+1)+C)#
Using #x=-2# gives us:
#6(-2)^2+13(-2)+6=A(-1)^2#
#A=4#
#6x^2+13x+6=4(x+1)^2+(x+2)(B(x+1)+C)#
Then using #x=-1# gives us:
#6(-1)^2+13(-1)+6=C#
#C=-1#
#6x^2+13x+6=4(x+1)^2+(x+2)(B(x+1)-1)#
Now using #x=0# (any value which hasn't been used can be used):
#6=4+2(B-1)#
#2(B-1)=2#
#B-1=1#
#B=2#
#6x^2+13x+6=4(x+1)^2+(x+2)(2(x+1)-1)#
#(6x^2+13x+6)/((x+2)(x+1)^2)=4/(x+2)+2/(x+1)-1/(x+1)^2#
#int4/(x+2)+2/(x+1)-1/(x+1)^2dx=4ln(abs(x+2))+2ln(abs(x+1))+int-1/(x+1)^2dx#
I left this one out so we can work on it separately.
We have #-(x+1)^-2#. We know that using the chain rule gives us #d/dx[f(x)^n]=nf(x)^(n-1)f'(x)#. We just have #-(x+1)^-2#, so #f(x)# must be #(x+1)^-1#
#d/dx[(x+1)^-1]=-(x+1)^-2#
#int4/(x+2)+2/(x+1)-1/(x+1)^2dx=4ln(abs(x+2))+2ln(abs(x+1))+(x+1)^-1+C#
