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#