`int 1/(x^2-9) dx` Use partial fractions to find the indefinite integral

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

To solve using the partial fraction method, the first step is to factor the denominator of the integrand.

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

Then, express it as a sum of two fractions. 

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

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

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

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

Then, assign values to x in such a way that either (x+3) or (x-3) will be zero. So, plug-in x =...

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

To solve using the partial fraction method, the first step is to factor the denominator of the integrand.

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

Then, express it as a sum of two fractions. 

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

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

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

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

Then, assign values to x in such a way that either (x+3) or (x-3) will be zero. So, plug-in x = 3 to get the value of A.

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

`1=A*6+B*0`

`1=6A`

`1/6=A`

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

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

`1=A*0 + B*(-6)`

`1=-6B`

`-1/6=B`

So the partial fraction decomposition of the integrand is:

`int 1/(x^2-9)dx=int (1/(6(x-3)) -1/(6(x+3)))dx`

Then, express it as difference of two integrals.

`=int 1/(6(x-3))dx - int 1/(6(x+3))dx`

`=1/6 int 1/(x-3)dx - 1/6 int 1/(x+3)dx`

And, apply the integral formula `int 1/u du = ln|u|+C` .

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

Therefore, `int 1/(x^2-9)dx=1/6ln|x-3| -1/6ln|x+3|+C` .