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)#