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

Question

1467 views around the world

You can reuse this answer
Creative Commons License

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,

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

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

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

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

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

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