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प्रश्न
`int 1/sqrt(x^2 - 8x - 20) "d"x`
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उत्तर
Let I = `int 1/sqrt(x^2 - 8x - 20) "d"x`
= `int 1/sqrt(x^2 - 2.4x + 16 - 16 - 20) "d"x`
= `int ("d"x)/sqrt((x - 4)^2 - 36) "d"x`
= `int ("d"x)/sqrt((x - 4)^2 - 6^2) "d"x`
= `log|(x - 4) + sqrt((x - 4)^2 - 6^2)| + "c"`
∴ I = `log|(x - 4) + sqrt(x^2 - 8x - 20)| + "c"`
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