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प्रश्न
`int sqrt((9 + x)/(9 - x)) "d"x`
उत्तर
Let I = `int sqrt((9 + x)/(9 - x)) "d"x`
= `int sqrt((9 + x)/(9 - x) xx (9 + x)/(9 + x)) "d"x`
= `int (9 + x)/(sqrt((9)^2 -x^2)) "d"x`
= `int [9/sqrt((9)^2 - x^2) + x/sqrt((9)^2 - x^2)] "d"x`
= `9 int 1/sqrt((9)^2 - x^2) "d"x +int x/sqrt((9)^2 - x^2) "d"x`
= `9 sin^(-1) (x/9) + "I"_1`
In I1, put (9)2 − x2 = t
∴ – 2x dx = dt
∴ x dx = `-1/2 "dt"`
∴ I1 = `-1/2 int "dt"/sqrt("t")`
∴ = `-1/2 * (("t"^(1/2))/(1/2)) + "c"`
= `- sqrt(9^2 - x^2) + "c"`
∴ I = `9 sin^(-1)(x/9) - sqrt(81 - x^2) + "c"`
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