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

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Answer 1 · 2017-12-05T16:42:09

#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#