Question

Determine whether the improper integral converges or diverges, and find the value of each that converges.

∫^[infinity]_-[infinity] x/x^2+1 dx

Answer 1

Converges to 0

Explanation:

Given is an improper integral where a= -infinity, b = infinity and

f
`f(x) = (x)/(x^2+1)`

`\int\limits^a_b {f(x)} \, dx`

This resembles the above integral where a = -b

f(x) let us check whether odd or even or neither

`f(-x) = (-x)/(x^2+1)`=-f(x)

Since f(x) is odd function and integral is of the form -a to a by properties of integrals, this value is 0

Hence answer is 0