`int 2/(9x^2-1)`

To solve using partial fraction method, the denominator of the integrand should be factored.

`2/(9x^2-1) = 2/((3x-1)(3x+1))`

Then, express it as sum of fractions.

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

To solve for the values of A and B, multiply both sides by the LCD of the fractions present.

`(3x-1)(3x+1)*2/((3x-1)(3x+1))=(A/(3x-1)+B/(3x+1))*(3x-1)(3x+1)`

`2 = A(3x+1)...

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`int 2/(9x^2-1)`

To solve using partial fraction method, the denominator of the integrand should be factored.

`2/(9x^2-1) = 2/((3x-1)(3x+1))`

Then, express it as sum of fractions.

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

To solve for the values of A and B, multiply both sides by the LCD of the fractions present.

`(3x-1)(3x+1)*2/((3x-1)(3x+1))=(A/(3x-1)+B/(3x+1))*(3x-1)(3x+1)`

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

Then, assign values to x in which either (3x+1) or (3x-1) will become zero.

So plug-in x=1/3 to get the value of A.

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

`2=A(1+1) + B(1-1)`

`2=A(2) + B(0)`

`2=2A`

`1=A`

Also, plug-in `x=-1/3` to get the value of B.

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

`2=A(-1+1)+B(-1-1)`

`2=A(0) + B(-2)`

`2=-2B`

`-1=B`

So the partial fraction decomposition of the integrand is

`int 2/(9x^2-1)dx`

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

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

Then, express it as difference of two integrals.

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

To evaluate each integral, apply substitution method.

`u=3x-1`

`du=3dx`

`1/3du=dx`

`w=3x+1`

`dw=3dx`

`1/3dw=dx`

Expressing the two integrals in terms of u and w, it becomes:

`=int 1/u * 1/3du - int 1/w*1/3dw`

`=1/3 int 1/u du - 1/3int 1/w dw`

To take the integral of this, apply the formula `int 1/x dx = ln|x|+C` .

`=1/3ln|u| - 1/3ln|w| + C`

And, substitute back u=3x-1 and w=3x+1.

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

**Therefore, `int 2/(9x^2-1)=1/3ln|3x-1| -1/3ln|3x+1|+C` .**