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Question
Evaluate the following : `int_(-1)^(1) (x^3 + 2)/sqrt(x^2 + 4)*dx`
Solution
Let I = `int_(-1)^(1) (x^3 + 2)/sqrt(x^2 + 4)*dx`
= `int_(-1)^(1) [x^3/sqrt(x^2 + 4) + 2/sqrt(x^2 + 4)]*dx`
= `int_(-1)^(1) x^3/sqrt(x^2 + 4)*dx + 2 int (1)/sqrt(x^2 + 4)*dx`
= I1 + 2I2 ...(1)
Let f(x) = `x^3/sqrt(x^2 + 4)`
∴ f(– x) = `(-x)^3/sqrt((-x)^2 + 4)`
= `x^3/sqrt(x^2 + 4)`
= – f(x)
∴ f is an odd function.
∴ `int_(-1)^(1)*dx` = 0, i.e.
I1 = `int_(-1)^(1) = x^3/sqrt(x^2 + 4)*dx` = 0 ...(2)
∵ (– x)2 = x2
∴ `(1)/sqrt(x^2 + 4)` is an even function.
∴ `int_(-1)^(1) f(x)*dx = 2int_0^(1) f(x)*dx`
∴ I2 = `2int_0^(1) 1/sqrt(x^2 + 4)*dx`
= `2[log (x + sqrt(x^2 + 4))]_0^1`
= `2g(1 + sqrt(1 + 4)) - log(0 + sqrt(0 + 4))]`
= `2[log (sqrt(5) + 1) - log 2]`
= `2 log ((sqrt(5 + 1))/2)` ...(3
From (1), (2) and (3, we get
I = `0 + 2[2 log ((sqrt(5 + 1))/2)]`
= `4log ((sqrt(5) + 1)/2)`.
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