How do you find int(x^2+1) / ( x^2 + 6x -3) dx using partial fractions?
1 Answer
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
4sqrt(3)A = 22+12sqrt(3)
Multiplying both sides by
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
