How do you evaluate the integral of #int (x-1)/(x^2+3x+2) dx# from 0 to 1?

#u=x^2+3x+2" "d u=(2x+3)d x#
#x-1=1/2(2x-2+3-3)" "x-1=1/2(2x+3-5)#
#=1/2int (2x+3-5)/(x^2+3x+2)d x" "#
#=1/2(int( (2x+3)d x)/(x^2+3x+2)-int (5d x)/(x^2+3x+2))#
#=1/2(int( d u)/u)-5int (d x)/(x^2+3x+2)#
#1/(x^2+3x+2)=A/(x+2)+B/(x+1)#
#1=A(x+1)+B(x+2)#
#x=-1 " "B=1" "x=-2" "A=-1#
#1/2[l n u-5(-int (d x)/(x+2)+int (d x)/(x+1) )]#
#1/2[l n u-5(l n(x+2)+l n(x+1))]_0^1+C#
#1/2[l n(x^2+3x+2)-5l n(x+2)-5(l n x+1)]_0^1+C#
#1/2[(l n 7-5l n 3-5l n2)-(l n 2-5l n2-5l n1)]+C#
#1/2[(l n7-5l n3-5l n2)-(-4l n2-5l n1)]" "l n1=0#
#1/2[l n7-5l n3-5l n2+4l n2]" "=1/2[l n7-5ln 3-l n2]#
#int _0^1 ((x-1)d x)/(x^2+3x+2)=-2,12014923742+C#

#(x-1)/(x^2+3x+2) = (x-1)/((x+2)(x+1))#

#A/(x+1)+B/(x+2) = (x-1)/((x+2)(x+1))#

#A(x+2)+B(x+1)=x-1#

#Ax+2A+Bx+B = x-1#

#(A+B)x + (2A+B) = 1x+(-1)#

Solve:
#{(A+B=1),(2A+B=-1):}#

#A=-2# and #B=3#

So
#int_0^1 (x-1)/(x^2+3x+2) dx = int_0^1 (-2/(x+1)+3/(x+2) )dx#

# = -2ln(x+1) + 3 ln(x+2)]_0^1#

# = [-2ln(2)+3ln(3)]-[-2ln(1)+3ln(2)]#

# = -5ln(2)+3ln(3)#

# = ln(27/32)#

Rewrite or approximate as you need.