Show that∫ ∞−∞2xdxx2 + 1 diverges even though limb→∞∫ b−b2xdxx2 + 1 = 0

To show that ∫ ∞−∞ 2x / (x^2 + 1) dx diverges, we can use the comparison test.

First, let's consider the integral ∫ ∞−∞ 2x / (x^2 + 1) dx. We can rewrite this integral as ∫ ∞−∞ 2x / (x^2 + 1) dx = ∫ ∞−∞ 2x / (x^2 + 1) dx.

Now, let's consider the integral ∫ b−b 2x / (x^2 + 1) dx, where b is a positive real number. We can rewrite this integral as ∫ b−b 2x / (x^2 + 1) dx = ∫ b−b 2x / (x^2 + 1) dx.

Next, let's find the antiderivative of 2x / (x^2 + 1). We can use the substitution method by letting u = x^2 + 1. Then, du = 2x dx. Rearranging, we have dx = du / (2x). Substituting back into the integral, we get ∫ 2x / (x^2 + 1) dx = ∫ du / u.

Integrating ∫ du / u, we get ln|u| + C, where C is the constant of integration. Substituting back u = x^2 + 1, we have ln|x^2 + 1| + C.

Now, let's evaluate the definite integral ∫ b−b 2x / (x^2 + 1) dx. Plugging in the limits of integration, we get ln|(b^2 + 1)| - ln|(b^2 + 1)| = ln|(b^2 + 1)/(b^2 + 1)| = ln|1| = 0.

Taking the limit as b approaches infinity, we have lim b→∞ ∫ b−b 2x / (x^2 + 1) dx = 0.

However, even though the limit of the definite integral is 0, the original integral ∫ ∞−∞ 2x / (x^2 + 1) dx still diverges. This is because the function 2x / (x^2 + 1) does not satisfy the conditions of the convergence test, such as the integral test or the comparison test.

Therefore, we have shown that ∫ ∞−∞ 2x / (x^2 + 1) dx diverges, even though lim b→∞ ∫ b−b 2x / (x^2 + 1) dx = 0.