How do you integrate #(5x+1)/((x+3)(x+2)(x-4))# using partial fractions?

Question

1554 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-04T19:22:09

#int ((5x+1)*dx)/[(x+3)(x+2)(x-4)]#

=#1/2*ln(x-4)+3/2*ln(x+2)-2ln(x+3)+C#

I decomposed integrand into basic fractions,

#(5x+1)/[(x+3)(x+2)(x-4)]#

=#A/(x+3)+B/(x+2)+C/(x-4)#

After expanding denominator,

#A(x+2)(x-4)+B(x+3)(x-4)+C(x+3)(x+2)=5x+1#

After setting #x=-3#, #7A=-14#, so #A=-2#

After setting #x=-2#, #-6B=-9#, so #B=3/2#

After setting #x=4#, #42C=21#, so #C=1/2#

Thus,

#int ((5x+1)*dx)/[(x+3)(x+2)(x-4)]#

=#-2*int (dx)/(x+3)+3/2*int (dx)/(x+2)+1/2*int dx/(x-4)#

=#1/2*ln(x-4)+3/2*ln(x+2)-2ln(x+3)+C#