How do you integrate #int (1-2x^2)/((x+1)(x+6)(x-7)) # using partial fractions?

The denominator has already been factored into distinct linear factors for us, so to construct a partial fraction decomposition we just need to find #A#, #B# and #C# to satisfy:

#(1-2x^2)/((x+1)(x+6)(x-7)) = A/(x+1) + B/(x+6) + C/(x-7)#

#=(A(x+6)(x-7)+B(x+1)(x-7)+C(x+1)(x+6))/((x+1)(x+6)(x-7))#

#=(A(x^2-x-42)+B(x^2-6x-7)+C(x^2+7x+6))/((x+1)(x+6)(x-7))#

#=((A+B+C)x^2+(-A-6B+7C)x+(-42A-7B+6C))/((x+1)(x+6)(x-7))#

Equating coefficients we get the system of linear equations:

#{ (A+B+C=-2), (-A-6B+7C=0), (-42A-7B+6C=1) :}#

Write this out as a matrix:

#((1, 1, 1, -2), (-1, -6, 7, 0), (-42, -7, 6, 1))#

Perform a sequence of row operations to make the left hand #3xx3# square into the identity matrix. Then the rightmost colum will give us the values of #A#, #B# and #C#.

Add row #1# to row #2# to get:

#((1, 1, 1, -2), (0, -5, 8, -2), (-42, -7, 6, 1))#

Add #42 xx# row #1# to row #3# to get:

#((1, 1, 1, -2), (0, -5, 8, -2), (0, 35, 48, -83))#

Add #7 xx# row #2# to row #3# to get:

#((1, 1, 1, -2), (0, -5, 8, -2), (0, 0, 104, -97))#

Multiply row #2# by #-1/5# to get:

#((1, 1, 1, -2), (0, 1, -8/5, 2/5), (0, 0, 104, -97))#

Subtract row #2# from row #1# to get:

#((1, 0, 13/5, -12/5), (0, 1, -8/5, 2/5), (0, 0, 104, -97))#

Divide row #3# by #104# to get:

#((1, 0, 13/5, -12/5), (0, 1, -8/5, 2/5), (0, 0, 1, -97/104))#

Subtract #13/5 xx# row #3# from row #1# to get:

#((1, 0, 0, 1/40), (0, 1, -8/5, 2/5), (0, 0, 1, -97/104))#

Add #8/5 xx# row #3# to row #2# to get:

#((1, 0, 0, 1/40), (0, 1, 0, -71/65), (0, 0, 1, -97/104))#

So:

#{ (A = 1/40), (B=-71/65), (C=-97/104) :}#

Then using #int 1/t dt = ln abs(t) + C# we have:

#int (1-2x^2)/((x+1)(x+6)(x-7)) dx#

#=int A/(x+1) + B/(x+6) + C/(x-7) dx#

#=1/40 ln abs(x+1) -71/65 ln abs(x+6) -97/104 ln abs(x-7) + c#