Find $\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}\:dx$ Find $$I=\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}dx$$
I have a very small query. I used the substitution $x+\frac{1}{x}=v$, then the limits of integration are both infinity. Can we say the integral is zero?
Instead what i did is:
after using that substitution we get:
$$I=\frac{1}{2} \lim _{x \rightarrow \infty} \ln \left|\frac{x^2-x+1}{x^2+x+1}\right|-\frac{1}{2} \lim _{x \rightarrow 0} \ln \left|\frac{x^2-x+1}{x^2+x+1}\right|=0$$
Is the second one better?
 A: If $I$ exists, then $I = I_1 + I_2$ where $$I_1 = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx, \quad f(x) = \frac{1-x^{-2}}{(x+x^{-1})^2-1}.$$
Since the antiderivative of $f$ is
$$F(x) = \int f(x) \, dx = \int \frac{dv}{v^2-1} = \frac{1}{2} \int \frac{1}{v-1} - \frac{1}{v+1} \, du = \frac{1}{2} \log \left|\frac{x-1+x^{-1}}{x+1-x^{-1}}\right| + C,$$ we have
$$I_1 = F(1) - \lim_{x \to 0^+} F(x) = -\frac{\log 3}{2} - 0 = -\frac{\log 3}{2},$$ and
$$I_2 = \lim_{x \to \infty} F(x) - F(1) = 0 + \frac{\log 3}{2}.$$
Therefore, $I = I_1 + I_2 = 0$.
Alternatively,  observe that the substitution $$x = 1/u, \quad dx = -u^{-2} \, du$$ gives
$$\begin{align}I 
&= \int_{x=0}^\infty f(x) \, dx \\
&= \int_{u=\infty}^0 -\frac{f(1/u)}{u^2} \, du \\
&= \int_{u=0}^\infty \frac{1-u^2}{u^2 ((u^{-1} + u)^2 - 1)} \, du \\
&= \int_{u=0}^\infty \frac{-(1-u^{-2})}{(u + u^{-1})^2 - 1} \, du \\
&= -I. \end{align}$$  Therefore $I = 0$.
This same idea can also be used to show $I_1 = -I_2$, rather than computing their values explicitly.
