Answer:
Converges to 0
Step-by-step explanation:
Given is an improper integral where a= -infinity, b = infinity and
f[tex]f(x) = \frac{x}{x^2+1}[/tex]
[tex]\int\limits^a_b {f(x)} \, dx[/tex]
This resembles the above integral where a = -b
f(x) let us check whether odd or even or neither
[tex]f(-x) = \frac{-x}{x^2+1}[/tex]=-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
