Question

Evaluate the following : `int_(-1)^(1) (1 + x^3)/(9 - x^2)*dx`

Answer 1

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`.