How do you find int(x^2+1) / ( x^2 + 6x -3) dx∫x2+1x2+6x−3dx using partial fractions?
1 Answer
int (x^2+1)/(x^2+6x-3) dx∫x2+1x2+6x−3dx
= x + ((18+11sqrt(3))/6)ln abs(x+3-2sqrt(3)) - ((54+11sqrt(3))/6)ln abs(x+3+2sqrt(3)) + C=x+(18+11√36)ln∣∣x+3−2√3∣∣−(54+11√36)ln∣∣x+3+2√3∣∣+C
Explanation:
x^2+6x-3 = (x+3)^2-12x2+6x−3=(x+3)2−12
= (x+3)^2-(2sqrt(3))^2=(x+3)2−(2√3)2
= (x+3-2sqrt(3))(x+3+2sqrt(3))=(x+3−2√3)(x+3+2√3)
(x^2+1)/(x^2+6x-3) = (x^2+6x-3-6x+4)/(x^2+6x-3)x2+1x2+6x−3=x2+6x−3−6x+4x2+6x−3
=1 + (-6x+4)/(x^2+6x-3)=1+−6x+4x2+6x−3
=1 + A/(x+3-2sqrt(3)) + B/(x+3+2sqrt(3))=1+Ax+3−2√3+Bx+3+2√3
=1 + (A(x+3+2sqrt(3)) + B(x+3-2sqrt(3)))/(x^2+6x-3)=1+A(x+3+2√3)+B(x+3−2√3)x2+6x−3
=1 + ((A+B)x + ((3+2sqrt(3))A + (3-2sqrt(3))B))/(x^2+6x-3)=1+(A+B)x+((3+2√3)A+(3−2√3)B)x2+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
