Question

How do you find `int(x^2+1) / ( x^2 + 6x -3) dx` using partial fractions?

Answer 1

`int (x^2+1)/(x^2+6x-3) dx`

`= x + ((18+11sqrt(3))/6)ln abs(x+3-2sqrt(3)) - ((54+11sqrt(3))/6)ln abs(x+3+2sqrt(3)) + C`

Explanation:

`x^2+6x-3 = (x+3)^2-12`

`= (x+3)^2-(2sqrt(3))^2`

`= (x+3-2sqrt(3))(x+3+2sqrt(3))`

`(x^2+1)/(x^2+6x-3) = (x^2+6x-3-6x+4)/(x^2+6x-3)`

`=1 + (-6x+4)/(x^2+6x-3)`

`=1 + A/(x+3-2sqrt(3)) + B/(x+3+2sqrt(3))`

`=1 + (A(x+3+2sqrt(3)) + B(x+3-2sqrt(3)))/(x^2+6x-3)`

`=1 + ((A+B)x + ((3+2sqrt(3))A + (3-2sqrt(3))B))/(x^2+6x-3)`

Equating coefficients, we have:

`{ (A+B = -6), ((3+2sqrt(3))A + (3-2sqrt(3))B = 4) :}`

From the first equation:

`B = -A-6`

Substituting in the second equation:

`(3+2sqrt(3))A+(3-2sqrt(3))(-A-6) = 4`

Which simplifies to:

`4sqrt(3)A-18+12sqrt(3) = 4`

Adding `18-12sqrt(3)` to both sides we get:

`4sqrt(3)A = 22+12sqrt(3)`

Multiplying both sides by `sqrt(3)/12` we find:

`A = (11sqrt(3))/6+3 = (18+11sqrt(3))/6`

Then:

`B = -A-6 = -(11sqrt(3))/6-9 = -(54+11sqrt(3))/6`

So:

`int (x^2+1)/(x^2+6x-3) dx`

`= int 1 + ((18+11sqrt(3))/6)(1/(x+3-2sqrt(3))) - ((54+11sqrt(3))/6)(1/(x+3+2sqrt(3))) dx`

`= x + ((18+11sqrt(3))/6)ln abs(x+3-2sqrt(3)) - ((54+11sqrt(3))/6)ln abs(x+3+2sqrt(3)) + C`