How do you integrate #f(x)=(x^2-2x)/((x^2-3)(x-3)(x-8))# using partial fractions?

Question

Christopher Allers · Accepted Answer

#int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]#

=#5/122Ln(x-8)#+#(21+5sqrt3)/732ln(x+sqrt3)#-#1/10Ln(x-3)#-#(21-5sqrt3)/6ln(x-sqrt3)+C# I decomposed into basic fractions, #(x^2-2x)/[(x^2-3)(x-3)(x-8)]# =#(Ax+B)/(x^2-3)+C/(x-3)+D/(x-8)# After expanding denominator, #(Ax+B)(x-3)(x-8)+C(x^2-3)(x-8)+D(x^2-3)(x-3)=x^2-2x# Set #x=3#, #-30C=3#, so #C=-1/10# Set #x=8#, #305D=48#, so #D=48/305# Set #x=0#, #24B+24C+9D=0#, so #B=-C-3/8*D=5/122# Set #x=2#, #14A+14B+14C+4D=-1#, so #A=-7/122# Hence, #int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]# =#-1/122int ((7x-5)*dx)/(x^2-3)-1/10int (dx)/(x-3)+5/122int (dx)/(x-8)# =#5/122Ln(x-8)-1/10Ln(x-3)+C-1/122*int ((7x-5)*dx)/(x^2-3)# Now I solved last integral, #int ((7x-5)*dx)/(x^2-3)# =#int ((7x-5)*dx)/[(x+sqrt3)(x-sqrt3)]# =#(21-5sqrt3)/6int (dx)/(x-sqrt3)-(21+5sqrt3)/6int (dx)/(x+sqrt3)# =#(21-5sqrt3)/6ln(x-sqrt3)-(21+5sqrt3)/6ln(x+sqrt3)# Thus, #int (x^2-2x)/[(x^2-3)(x-3)(x-8)*dx]# =#5/122Ln(x-8)+(21+5sqrt3)/732ln(x+sqrt3)-1/10Ln(x-3)-(21-5sqrt3)/6ln(x-sqrt3)+C#

William Abdalla · Answer

undefined To integrate the function $ f(x) = \frac{{x^2 - 2x}}{{(x^2 - 3)(x - 3)(x - 8)}} $ using partial fractions, follow these steps:

1. Factor the denominator $ (x^2 - 3)(x - 3)(x - 8) $ into linear and irreducible quadratic factors.
2. Express $ f(x) $ as the sum of partial fractions with undetermined coefficients.
3. Find the values of the coefficients by equating the numerators of the partial fractions with the original function.
4. Integrate each partial fraction separately.
5. Combine the results to get the final integrated expression.

After performing the steps, you will have the integrated expression.