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प्रश्न
Prove that:
`int 1/(a^2 - x^2) dx = 1/2 a log ((a +x)/(a-x)) + c`
उत्तर
Let I = `int1/(a^2 - x^2)dx`
= `intdx/((a - x)(a + x))`
= `1/(2a) int (1/(a + x) + 1/(a - x))dx`
= `1/(2a) [int dx/(a + x) + int dx/(a - x)]`
= `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`
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