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int1/(x^2-9)dx...

`int1/(x^2-9)dx`

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` int(1)/(x^(2)-9)dx = int (1)/(x^(2)-3^(2))dx`
`= int (1)/((x+3)(x-3))dx= I "(say)"`
`" Let " (1)/((x+3)(x-3))=(A)/((x+3))+(B)/((x-3))`
` rArr 1=A (x-3) +B(x+3)`
`x=3"0 then " 1=0+ B(3+3) rArr B=(1)/(6)`
`x=-3 " then " 1=A (-3-3) +0 rArr A=-(1)/(6)`
`:. I= int (1)/((x+3)(x-3))dx`
`=int((-1))/(6(x+3))dx+ int (1)/(6(x-3))dx`
`=-(1)/(6)log |x+3|+(1)/(6)log |x-3|+c`
`=(1)/(6)log |(x-3)/(x+3)|+c`
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