`int(3-x)/(3x^2-2x-1)dx`

Let's use partial fraction decomposition on the integrand,

`(3-x)/(3x^2-2x-1)=(3-x)/(3x^2+x-3x-1)`

`=(3-x)/(x(3x+1)-1(3x+1))`

`=(3-x)/((3x+1)(x-1))`

Now form the partial fractions using the denominator,

`(3-x)/((3x+1)(x-1))=A/(3x+1)+B/(x-1)`

Multiply equation by the denominator `(3x+1)(x-1)`

`=>(3-x)=A(x-1)+B(3x+1)`

`=>3-x=Ax-A+3Bx+B`

`=>3-x=(A+3B)x+(-A+B)`

comparing the coefficients of the like terms,

`A+3B=-1`   ----------------(1)

`-A+B=3`      ----------------(2)

Now let's solve the above equations to get A and B,

Add the equations 1 and 2,

`4B=-1+3`

`4B=2`

`B=2/4`

`B=1/2`

Plug in the value of B in equation 1,

`A+3(1/2)=-1`

`A+3/2=-1`

`A=-1-3/2`

`A=-5/2`

Plug in the value of A and B in the partial fraction template,

`=(-5/2)/(3x+1)+(1/2)/(x-1)`

`=-5/(2(3x+1))+1/(2(x-1))`

So, `int(3-x)/(3x^2-2x-1)dx=int(-5/(2(3x+1))+1/(2(x-1)))dx`

Apply the sum rule,

`=int-5/(2(3x+1))dx+int1/(2(x-1))dx`

Take the constant out,

`=-5/2int1/(3x+1)dx+1/2int1/(x-1)dx`

Now let's evaluate both the above integrals separately,

`int1/(3x+1)dx`

Apply integral substitution:`u=3x+1`

`=>du=3dx`

`=int1/u(du)/3`

Take the constant out,

`=1/3int1/udu`

Use the common integral:`int1/xdx=ln|x|`

`=1/3ln|u|`

Substitute back `u=3x+1`

`=1/3ln|3x+1|`

Now evaluate the second integral.

`int1/(x-1)dx`

Apply integral substitution: `u=x-1`

`du=1dx`

`=int1/udu`

Use the common integral:`int1/xdx=ln|x|`

`=ln|u|`

Substitute back `u=x-1`

`=ln|x-1|`

`int(3-x)/(3x^2-2x-1)dx=-5/2(1/3ln|3x+1|)+1/2ln|x-1|` 

Simplify and add a constant C to the solution,

`=-5/6ln|3x+1|+1/2ln|x-1|+C`