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

1 Answer
Nov 15, 2017

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

Explanation:

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