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int 1/ a^2 - x^2 dx = 1 / 2a log ( a + x...

`int 1/ a^2 - x^2 dx = 1 / 2a log ( a + x / a - x ) +c`

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`int(1)/(a^(2)-x^(2))dx=int(1)/((a-x)(a+x))dx`
`=(1)/(2a) int((a-x)+(a+x))/((a-x)(a+x))dx`
`=(1)/(2a) int((1)/(a+x)+(1)/(a-x))dx`
`=(1)/(2a)[log|a+x|+(log|a-x|)/(-1)]+c`
`=(1)/(2a)[log|a+x|-log|a-x|]+c`
`=(1)/(2a)log|(a+x)/(a-x)|+c`
`"Hence", int (1)/(a^(2)-x^(2))dx=(1)/(2a)log|(a+x)/(a-x)|+c`
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