You have a second-degree factor and a first-degree factor in the denominator. #x^2 + 1# can't be factored any further as all real numbers, so this will separate into:
#A/(x - 1) + (Bx+C)/(x^2 + 1)#
Cross-multiply:
#= (A(x^2 + 1) + (Bx+C)(x-1))/((x-1)(x^2+1))#
Multiply it out (and arrange it to look descending if you want):
#= (Ax^2 + A + Bx^2 - Bx + Cx - C)/((x-1)(x^2+1))#
However you do it, you should see now that there are #x^2#, #x#, and #x^0# (constant) terms.
#= (color(highlight)((A+B))x^2 + color(highlight)((-B + C))x + color(highlight)((A - C)))/((x-1)(x^2+1))#
I purposefully changed #- (B - C)# to #+ (-B + C)# to match the original numerator's adding #x#. We can equate them back to the original equation's numerator terms' coefficients.
1) #A + B = 1#
2) #-B + C = 1#
3) #A - C = -6#
Now, here's what I would do. The parentheses indicate which equation was used when. Single parentheses indicate newly-numbered equations.
#A = 1 - B# (1)
#1 - B - C = -6# (1#->#3)
4) #B = 7 - C#
#C - 7 + C = 1# (4 #-># 2)
#2C = 8#
#color(green)(C = 4)#
#A - 4 = -6# (3)
#color(green)(A = -2)#
#-2 + B = 1 -> color(green)(B = 3)# (1)
Plug them back in and integrate:
#=> int-2/(x-1)dx + int(3x + 4)/(x^2 + 1)dx#
Notice how you can separate the second integral (some people may have trouble thinking of that):
#= -2int1/(x-1)dx + int(3x)/(x^2 + 1)dx + 4int 1/(x^2 + 1)dx#
And prepare the now-second integral for u-substitution by multiplying by #3/2*2/3# since #2/3*3x = 2x# and #d(x^2) = 2xdx#:
#= -2int1/(x-1)dx + 3/2int(2x)/(x^2 + 1)dx + 4int 1/(x^2 + 1)dx#
For the first integral, it is #-2ln|x-1|#. With the second one, letting #u = x^2+1#, #du = 2xdx# (that's what we were preparing for) and thus it becomes:
#3/2int1/udu#
so it is therefore #3/2ln|x^2+1|#. The third is #4arctanx# since #4d/(dx)[arctanx] = 4*1/(1+x^2)#.
#=> color(blue)(-2ln|x-1| + 3/2ln|x^2+1| + 4arctanx + C)#
