How do you integrate #int (x-9)/((x+3)(x-7)(x-5)) # using partial fractions?

As the other answer was unfortunately deleted, I will answer this question.

To compute a partial fractions decomposition means to find #A#, #B# and #C# such as

#(x-9)/((x+3)(x-7)(x-5)) = A / (x+3) + B / (x-7) + C / (x-5)#

... multiply both sides with the denominator #(x+3)(x-7)(x-5)#...

# <=>" " x - 9 = A(x-7)(x-5) + B(x+3)(x-5) + C (x+3)(x-7)#

# <=>" " x - 9 = A(x^2 - 12x + 35) + B(x^2 - 2x - 15) + C (x^2 - 4x - 21)#

# <=>" " x - 9 = A * x^2 - A * 12x + A * 35 + B * x^2 - B * 2x - B * 15 + C * x^2 - C * 4x - C * 21#

... identify the #color(red)(x^2)# terms, the #color(blue)(x)# terms and the #"constant"# terms:

# <=>" " color(red)(0 * x^2) + color(blue)(x) - 9 = color(red)(A * x^2) color(blue)(- 12A * x) + 35A + color(red)(B * x^2) color(blue)(- 2B * x) - 15B + color(red)(C * x^2) color(blue)(- 4C * x) - 21C#

This leads us to the following linear equation system:

#{ (" " (I) " "0 = " "A + B " "+C " " color(red)(x^2) " terms"), (" " (II) " "1 = -12A - 2 B " "- 4C " " color(blue)(x) " terms"), (" " (III) -9 = " "35A - 15B - 21C " constant terms") :}#

Now, let's solve this linear equation system.

To eliminate #B#, let's compute #(IV) = 2 * (I) + (II)# and #(V) = 15 * (I) + (III)#:

#(IV) " " 1 = -10 A - 2C#
#(V) " " -9 = " "50 A - 6C#

Now, we can compute #(VI) = 5 * (IV) + (V)# to eliminate #A#:

#(VI)" " -4 = -16 C#

Thus, we can solve #(VI)# for #C# and find #C = 1/4#.

Plugging #C# into #(IV)# gives us #A = -3/20# and finally, plugging #A# and #C# into #(I)# gives us #B = - 1/10#.

Now, we can perform the partial fraction decomposition and solve the integral:

# int (x-9)/((x+3)(x-7)(x-5)) "d"x#

#= int [-3/20 * 1 / (x+3) -1/10 * 1/ (x-7) + 1/4 * 1/ (x-5) ] "d"x#

#= -3/20 int [ 1 / (x+3)"d"x -1/10 int 1 / (x-7) "d"x + 1/4 int 1/ (x-5) "d"x#

#= -3/20 ln abs (x+3) -1/10 ln abs (x-7) + 1/4 ln abs (x-5) + C#