Question

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

Answer 1

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

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

Explanation:

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`