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Question
Evaluate the following : `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`
Solution
Let I = `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`
= `int_(-1)^(1)[1/(9 - x^2) + x^3/(9 - x^2)]*dx`
= `int_(-1)^(1) 1/(9 - x^2)*dx + int_(-1)^(1) x^3/(9 - x^2)*dx`
∴ I = I1 + I2 ....(1)
I1 = `int_(-1)^(1) 1/(3^2 - x^2)*dx`
= `(1)/(2 xx 3)[log|(3 + x)/(3 - x)|]_(-1)^(1)`
= `(1)/(6)[log (4/2) - log(2/4)]`
= `(1)/(6)[log(2/(1/2))]`
= `(1)/(6)log 4`
= `(1)/(6)log 2^2`
= `(1)/(6) xx 2log2`
= `(1)/(3)log2` ...(2)
I2 = `int_(-1)^(1) x^3/(9 - x^2)*dx`
Let f(x) = `x^3/(9 - x^2)`
∴ f(– x) = `(- x)^3/(9 - (- x)^2`
= `(-x)^3/(9 - x^2)`
= – f(x)
∴ f is an odd function.
∴ `int_(-1)^(1) f(x)*dx` = 0
∴ I2 = `int_(-1)^(1) x^3/(9 - x^2)*dx` = 0 ...(3)
From (1),(2) and (3), we get
I = `(1)/(3)log2 + 0`
= `(1)/(3)log2`.
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