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

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`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. 


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


`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)`




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


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



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` .

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