I'm not sure sure whether Lebesgue integration helps to solve that integral, though I think it doesn't. Perhaps a little complex analysis?
$$f(z)=\frac{1+z}{1+z^2}=\frac{1+z}{(z+i)(z-i)}$$
and we can take the contour determined by
$$C_R:=[-R,R]\cup\gamma_R:=\{z\in\Bbb C\;;\;|z|=R\;,\;\text{Im}(z)\ge 0\}\;,\;\;R>>0$$
Within the domain bounded by the above curve we've only one simple pole, so
$$Res_{z=i}(f)=\lim_{z\to i}(z-i)f(z)\stackrel{\text{l'Hospital}}=\frac{1+i}{2i}$$
so by Cauchy's Residue Theorem
$$2\pi i\cdot\frac{1+i}{2i}=\oint\limits_{C_R}f(z)dz+\int_{-R}^R\frac{1+x}{1+x^2}dx$$
and passing to the limit when $\,R\to\infty\,$ and taking the real parts we get
$$\pi=\int_{-\infty}^\infty\frac{x+1}{x^2+1}dx$$
getting btw that the imaginary part has also the same value...
It should be noted that the imaginary part is due entirely to the loop integral; the imaginary part of the integral along the real axis is zero, as expected.
Since this is a question about rigor we should state what integral we are using. From (1) This integral is not Lebesgue integrable as it is not absolutely convergent, it is an improper Riemann integral though so we will use the Riemann integral.
Let us have a visual of the function being integrated
When we state the purely real integral $$I_1 = \int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$ this is short-hand for $$I_1 = \lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} f(x) dx$$ where $$f(x) = \begin{cases}
\frac{\sin(x)}{x} & x \not = 0 \\
1 & x = 0 \\
\end{cases}$$
The convergence of this integral must be addressed. It does not converge absolutely so we should attempt to cancel parts of the integral with itself. We will show that the right side (from $2\pi$ to $\infty$) converges by chopping it up and subtracting the negative part of each sine wave from the positive part. For each segment we have
$$\begin{align}
&\, \left|\int_{\pi 2 k}^{\pi (2 k + 2)} \frac{\sin(x)}{x} dx\right| \\
\le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{x} - \frac{\sin(x)}{x + \pi} dx\right| \\
\le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{\pi (2 k + 1)} - \frac{\sin(x)}{\pi (2 k + 2)} dx\right| \\
\le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{\pi} \left[\frac{1}{2 k + 1} - \frac{1}{2 k + 2}\right] dx\right| \\
\end{align}$$
and the alternating harmonic series converges. Note that each segment is positive and their sum converges absolutely. Note that we showed the integral from $0$ to $\infty$ converges independently of the integral from $-\infty$ to $0$.
The finite part from $0$ to $2 \pi$ easily converges as we always have $\sin(x)/x \le 1$.
When we state a contour integral like
$$I_2 = \mathrm{p.v.} \int_\gamma \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz$$
($\gamma$ denoting a line from $-\infty$ to $\infty$).
the meaning (2) is that the we delete from the contour $\gamma$ an $\epsilon$ sized ball around the singularity and take the limit of $\epsilon$ to 0.
So $$I_2 = \lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz$$
Regarding convergence of $I_2$, this integral (or half of it, to be precise) drops out algebraically from an application of Cauchy's Theorem (that a contour integral around a closed curve not containing poles gives 0) to the meromorphic function $e^{iz}/z$. The details are here.
Now to show $I_1 = I_2$. The single exceptional point at $x=0$ has no bearing on the value of the integral $I_1$ so we may split $I_1$ around 0 and add a $\lim_{\epsilon \to 0}$ on it. Now $I_1 - I_2 = 0$ can be shown. Rewrite $I_2$ to use the complex $\sin$ function.
$$\begin{align}
I_1 - I_2 =& \left[\lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} f(x) dx \right] - \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz\right] \\
=& \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{\sin(x)}{x} dx \right] - \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{\sin(z)}{z} dz\right] \\
=& \lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \left[\frac{\sin(x)}{x} - \frac{\sin(x)}{x} \right] dx \\
=& 0
\end{align}$$
Best Answer
The Cauchy principal value is very important, especially in cases where the Lebesgue integral (which it seems you refer to as the improper integral) does not exist. The issue is that the Lebesgue integral doesn't deal too well with really big oscillations. Indeed, a measurable function $f$ is Lebesgue integrable if and only if $|f|$ is, so oscillations don't matter - only the magnitude does. This becomes problematic when dealing with functions such as $\frac{\sin x}{x}$ on $(0,\infty)$ or $\frac{1}{x}$ on $(-1,1)\setminus\{0\}$.
A purpose of the Cauchy principal value is to rectify this problem, to take into account oscillations like the Riemann integral does and give a meaningful number that represents the integral (i.e. scaled average) of the function in question. The Cauchy p.v. of $\frac{1}{x}$ is $\lim\limits_{\epsilon \to 0^+} \int_{-1}^{-\epsilon} \frac{1}{x}dx+\int_{\epsilon}^1 \frac{1}{x}dx = 0$, which coincide with our intuition for what the average value of $\frac{1}{x}$ should be.
The most prominent use of the Cauchy principal value is the Hilbert transform, in which we study $\int \frac{f(x-y)}{y}dy$, which of course, needs to be defined properly. It is critical here that we don't just say the improper integral exists, but rather get a quantitative sense of the oscillation present.