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.
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.
Best Answer
If you expand $\frac{1}{\exp(-z)-1}$ in a Taylor series around $z=0$, you notice that $$ \frac{1}{\exp(-z)-1}\approx -\frac{1}{z}-\frac{1}{2}-\frac{z}{12}+\ldots $$ so the function has indeed a non-integrable singularity at $z=0$. One way to assign a finite result to your integral is to subtract and re-add the singular part to your integral as $$ \int_{-\infty}^{\infty}\dfrac{1}{(z+1)^2+4} \left[\dfrac{1}{\exp(-z)-1}+\frac{1}{z}-\frac{1}{z}\right]= $$ $$ \int_{-\infty}^{\infty}\dfrac{1}{(z+1)^2+4} \left[\dfrac{1}{\exp(-z)-1}+\frac{1}{z}\right]-\mathcal{P}\int_{-\infty}^{\infty}\dfrac{1}{(z+1)^2+4} \frac{1}{z}\ , $$ where $\mathcal{P}$ is Cauchy's principal part. The final result is $-0.383448...$ (checked and agreed with Claude's answer below.)