How do you integrate #int(x+1)/((2x-4)(x-4)(x-7))# using partial fractions?

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Answer 1 · 2017-12-16T18:56:41

#3/20Ln(Abs(x-2))-5/12(Abs(x-4))+4/15Ln(Abs(x-7))+C#

#(x+1)/[(2x-4)*(x-4)(x-7)]#

=#1/2*(x+1)/[(x-2)*(x-4)(x-7)]#

I decomposed #(x+1)/[(x-2)*(x-4)(x-7)]# into basic fractions,

#(x+1)/[(x-2)*(x-4)(x-7)]#

=#A/(x-2)+B/(x-4)+C/(x-7)#

After expanding denominator,

#A(x-4)(x-7)+B(x-2)(x-7)+C(x-2)(x-4)=x+1#

Set #x=2#, #10A=3#, so #A=3/10#

Set #x=4#, #-6B=5#, so #B=-5/6#

Set #x=7#, #15C=8#, so #C=8/15#

Thus,

#int (x+1)/[(2x-4)*(x-4)(x-7)]*dx#

=#1/2int (x+1)/[(2x-4)*(x-4)(x-7)]*dx#

=#1/2*3/10int (dx)/(x-2)-1/2*5/6int (dx)/(x-4)+1/2*8/15int (dx)/(x-7)#

=#3/20int (dx)/(x-2)-5/12int (dx)/(x-4)+4/15int (dx)/(x-7)#

=#3/20Ln(Abs(x-2))-5/12(Abs(x-4))+4/15Ln(Abs(x-7))+C#