The simplest way to do this kind of partial fraction integral is to employ the Heaviside Method of partial fraction decomposition.

Here's the idea: Our goal is to take the expression in the integral and split it into 3 separate fractions, each of which is easier to integrate on their own. This means we need to (in this example, at least) find 3 coefficients #A, B, and C# such that the following equation is true for every x:

#(3x-2x^2)/((x+9)(x+7)(x+1)) = A/(x+9) + B/(x+7) + C/(x+1)#

The Heaviside Method is often used to make this process much simpler. It essentially says you should take one factor of the denominator in the original expression at a time, set it equal to 0 to determine that #x# value (ie, find the root for that expression), then substitute it into the original fraction expression everywhere **except** that denominator factor. Whatever you come out with is the coefficient of that factor in the separated fraction. Let's see that in practice:

**Find A**

The denominator in the A term is #(x+9)#. If we set that equal to 0 and solve, we find #x = -9#. We now go to the original fraction in the integral and we substitue #x = -9# in for *every* term **except** the #(x+9)# factor in the denominator, and simplify (I'm drawing in that term just so you can see we're ignoring it):

#(3(-9)-2(-9)^2)/(color(gray)(cancel((x+9)))(-9+7)(-9+1)) = (-27-162)/((-2)(-8))=-189/16#

Thus, #A = -189/16#

**Find B**

Again, the denominator in the B term is #(x+7)#. Setting this equal to 0 and solving leads to #x = -7#. Going back to the original fraction and substituting:

#(3(-7)-2(-7)^2)/((-7+9)color(gray)(cancel((x+7)))(-7+1)) = (-21-98)/((2)(-6))=119/12#

Thus, #B = 119/12#

**Find C**

Lastly, the denominator in the C term is #(x+1)#. Setting this equal to 0 and solving leads to #x = -1#. Going back to the original fraction and substituting:

#(3(-1)-2(-1)^2)/((-1+9)(-1+7)color(gray)(cancel((x+1)))) = (-3-2)/((8)(6))=-5/48#

Thus, #C = -5/48#

**Replace Integral and Solve**

Now we have the expansion of fractions we need to solve the problem:

#int (3x-2x^2)/((x+9)(x+7)(x+1)) dx #

#=int (A/(x+9) + B/(x+7) + C/(x+1)) dx#

# = int ((-189/16)/(x+9) + (119/12)/(x+7) - (5/48)/(x+1) ) dx#

# = -189/16 int 1/(x+9) dx + 119/12 int 1/(x+7) dx - 5/48 int 1/(x+1) dx #

# = -189/16 ln |x+9| + 119/12 ln |x+7| - 5/48 ln |x+1| + C #