Answer
The given improper integral is:
$$
\int_{-\infty}^{\infty} \frac{d x}{4 x^{2}+4 x+5} =\frac{\pi}{4}
$$
Work Step by Step
$$
\begin{aligned} \int_{-\infty}^{\infty} \frac{d x}{4 x^{2}+4 x+5} &=\int_{-\infty}^{\infty} \frac{d x}{(2x+1)^{2}+4} \\
&=\int_{-\infty}^{\infty} \frac{\frac{1}{2} d u}{u^{2}+4}\\
& \quad\quad\quad\left[\text { Let } u=2x+1 \text { so that } du=2 d x \text {: } \right]\\
&=\frac{1}{2} \int_{-\infty}^{0} \frac{d u}{u^{2}+4}+\frac{1}{2} \int_{0}^{\infty} \frac{d u}{u^{2}+4} \\
&=\frac{1}{2} \lim _{t \rightarrow-\infty} \int_{t}^{0} \frac{d u}{u^{2}+4}+\frac{1}{2} \lim _{t \rightarrow \infty} \int_{0}^{t} \frac{d u}{u^{2}+4}\\
&=\frac{1}{2} \lim _{t \rightarrow-\infty}\left[\frac{1}{2} \tan ^{-1}\left(\frac{1}{2} u\right)\right]_{t}^{0}+ \\
& \quad\quad\ \quad\quad\ +\frac{1}{2} \lim _{t \rightarrow \infty}\left[\frac{1}{2} \tan ^{-1}\left(\frac{1}{2} u\right)\right]_{0}^{t} \\
&=\frac{1}{4}\left[0-\left(-\frac{\pi}{2}\right)\right]+\frac{1}{4}\left[\frac{\pi}{2}-0\right] \\
&=\frac{\pi}{4} \end{aligned}
$$
Thus the given improper integral is:
$$
\int_{-\infty}^{\infty} \frac{d x}{4 x^{2}+4 x+5} =\frac{\pi}{4}
$$
