Question

How do you integrate `f(x)=x/((x^2+2)(x+4)(x-7))` using partial fractions?

Answer 1

`int \ f(x) \ dx = -29/1700ln(x^2+1) - 3/850arctanx + 4/187ln|x+4| + 7/550ln|x-7| + C`

Explanation:

We have:

`f(x) = x/( (x^2+1) (x+4) (x-7) )`

We can decompose into partial fractions, which will be of the form:

`x/( (x^2+1) (x+4) (x-7) ) = (Ax+B)/(x^2+1) + C/(x+4) + D/(x-7)`

And after putting over a common denominator we end up with:

`x -= (Ax+B)(x+4)(x-7) + C(x^2+1)(x-7) + D(x^2+1)(x+4)`

Where the coefficients `A,B,C,D` are constants to be found.

We can use direct substitution

`x=7 => 7=0+0+D(49+1)(7+4) =>D=7/550`
`x=-4 => -4=0+C(16+1)(-4-7)+0=>C=4/187`

And we can compare coefficients:

`x^3: 0 = A+C+D => A=-29/850`
`x^0: 0=-28B-7C+4D => B=-3/850`

Thus we have:

`f(x) = (-29/850x-3/850)/(x^2+1) + (4/187)/(x+4) + (7/550)/(x-7)`

And so:

`int f(x) dx= -1/850int(29x+3)/(x^2+1) dx + 4/187int1/(x+4)dx + 7/550int1/(x-7) dx`
`" "= -1/850int(29x+3)/(x^2+1) dx + 4/187ln|x+4| + 7/550ln|x-7| + C`

We need further work for the first integral:

`int \ (29x+3)/(x^2+1) dx = int \ (29x)/(x^2+1) + 3/(x^2+1) dx`
`" " = 29/2int \ (2x)/(x^2+1) dx + 3int 1/(x^2+1) dx`
`" " = 29/2ln|x^2+1| + 3arctanx`

Combining we get:

`int \ f(x) \ dx = -1/850 {29/2ln|x^2+1| + 3arctanx} + 4/187ln|x+4| + 7/550ln|x-7| + C`

And, as `x^2+1 gt 0 AA x in RR` we have:

`int \ f(x) \ dx = -29/1700ln(x^2+1) - 3/850arctanx + 4/187ln|x+4| + 7/550ln|x-7| + C`