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

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Answer 1 · 2017-11-15T21:35:59

#2/17*ln(x^2-2)+(7sqrt(2))/68*Ln((x-sqrt2)/(x+sqrt2))-4/17*Ln(3x-1)+C#

I decomposed integrand into basic fractions,

#(x+1)/[(x^2-2)*(3x+1)]=(Ax+B)/(x^2-2)+C/(3x-1)#

After expanding denominator,

#(Ax+B)(3x-1)+C(x^2-2)=x+1#

Set #x=1/3#, #-17C/9=4/3#, so #C=-12/17#

Set #x=0#, #-B-2C=1#, so #B=-2C-1=7/17#

Set #x=1#, #2A+2B-C=2#, so #A=1/2*(2+C-2B)=4/17#

Hence,

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

=#1/17*int ((4x+7)*dx)/(x^2-2)-1/17*int (12*dx)/(3x-1)#

=#2/17*int (2x*dx)/(x^2-2)+(7sqrt(2))/68*int (2sqrt2*dx)/(x^2-2)-4/17*int (3*dx)/(3x-1)#

=#2/17*ln(x^2-2)+(7sqrt(2))/68*Ln((x-sqrt2)/(x+sqrt2))-4/17*Ln(3x-1)+C#