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प्रश्न
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
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उत्तर
Let I = `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
∴ I = `int_(-1)^1 1/(9 - x^2) "d"x + int_(-1)^1 x^3/(9 - x^2) "d"x`
= I1 + I2 ...(say) ......(i)
Let `"f"(x) = 1/(9 - x^2)`
∴ f(−x) = `1/(9 - (- x)^2`
= `1/(9 - x^2)`
= f(x)
∴ f(x) is an even function.
∴ I1 = `int_(-1)^1 1/(9 - x^2) "d"x`
= `2 int_0^1 1/(9 - x^2) "d"x`
= `2 int_0^1 1/(3^2 - x^2) "d"x`
= `2[1/(2 xx 3)* log|(3 + x)/(3 - x)|]_0^1`
= `1/3[log(4/2) - log(1)]`
∴ I1 = `1/3 log 2`
Let g(x) = `x^3/(9 - x^2)`
∴ g(−x) = `(-x)^3/(9 - (- x)^2`
= `(-x^3)/(9 - x^2)`
= − g(x)
∴ g (x) is an odd function.
∴ I2 = `int_(-1)^1 x^3/(9 - x^2) "d"x` = 0
From (i), we get
I = I1 + I2
∴ I = `1/3 log 2 + 0`
∴ I = `1/3 log 2`
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