Question

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

Answer 1

`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`