Question

Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2)  "d"x`

Answer 1

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`