First if we consider the integral:

\begin{equation}
I(a) = \int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx = \frac{\pi}{a}
\end{equation}

Then we can see:

\begin{equation}
I(a) + cI(b) = \left[\int_{-\infty}^{\infty}\frac{1}{x^2 + a^2}\:dx + c\int_{-\infty}^{\infty}\frac{1}{x^2 + b^2}\:dx \right] = \frac{\pi}{a} + c\frac{\pi}{b} = \pi\left[\frac{1}{a} + c\frac{1}{b} \right]
\end{equation}

We can see that this can be expanded to any number of $I(a)$ terms and any sequence of $c_i$ values, i.e.

\begin{equation}
\sum_{i = 1}^{\infty} (-1)^{i + 1} \frac{1}{i + 1}I(i) = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}

Thus,

\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}

Or as another example,

\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{1}{x^2 + i^4} \:dx = \frac{ \pi^3}{6}
\end{equation}

What's even better (IMO) is that we can take each integral spoken to and apply (through Glasser Master Theorem)

$$x = t - \sum_{i = 1}^{n - 1}\frac{\left| d_{i}\right|}{ t - e_{i}}$$

Where $d_i, e_i \in \mathbb{R}$ and $n \in N$ that the value of the integrals remain unchanged!!

As an addendum to the example:

\begin{equation}
\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \pi \sum_{i = 1}^{\infty} \frac{(-1)^{i + 1}}{i + 1} = \pi\ln(2)
\end{equation}

If we call on this integral as addressed here we see that

\begin{equation}
2\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = 2\sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx =\pi\ln(2)
\end{equation}

Or,
\begin{equation}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\tan(x)} \:dx = \sum_{i = 1}^{\infty} \int_{-\infty}^{\infty} \frac{(-1)^{i + 1}}{x^2 + \left(i + 1\right)^2} \:dx = \frac{\pi\ln(2)}{2}
\end{equation}