Answer
\[\frac{π}{4}\]
Work Step by Step
Let \[I=\int_{-\infty}^{\infty}\frac{dx}{4x^2+4x+5}\]
\[I=\lim_{t\rightarrow \infty}\int_{-t}^{t}\frac{dx}{4x^2+4x+5}\;\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{dx}{4x^2+4x+5}\]
\[I_1=\frac{1}{4}\int\frac{dx}{x^2+x+\frac{5}{4}}\]
\[I_1=\frac{1}{4}\int\frac{dx}{x^2+x+\frac{5}{4}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2}\]
\[I_1=\frac{1}{4}\int\frac{dx}{\left(x+\frac{1}{2}\right)^2+1^2}\]
\[\left[\int\frac{dx}{a^2+x^2}=\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\right]\]
\[I_1=\frac{1}{4}\tan^{-1}\left(x+\frac{1}{2}\right)+C\;\;\;\;\ldots (2)\]
Where $C$ is constant of integration
Using (2) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\frac{1}{4}\tan^{-1}\left(x+\frac{1}{2}\right)\right]_{-t}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\frac{1}{4}\tan^{-1}\left(t+\frac{1}{2}\right)-\frac{1}{4}\tan^{-1}\left(-t+\frac{1}{2}\right)\right]\]
\[I=\frac{1}{4}\left(\frac{π}{2}\right)-\frac{1}{4}\left(\frac{-π}{2}\right)=\frac{2π}{8}\]
\[I=\frac{π}{4}\]
Hence \[I=\frac{π}{4}\;\;\;.\]
