## Calculus (3rd Edition)

$\pi$
We will compute the integral as follows: $\int_{-\infty}^{\infty} \dfrac{dx}{1+x^2}=\int_{-\infty}^0 \dfrac{dx}{1+x^2}+\int_{0}^{\infty} \dfrac{dx}{1+x^2}\\=\lim\limits_{R \to \infty}\int_{-R}^0 \dfrac{dx}{1+x^2}+ \lim\limits_{R \to \infty}\int_{0}^{R} \dfrac{dx}{1+x^2}\\=\lim\limits_{R \to \infty} \arctan (x)_{-R}^0+\lim\limits_{R \to \infty} \arctan (x)_{0}^R\\=\lim\limits_{R \to \infty} [\arctan (0)-\arctan (-R)]+\lim\limits_{R \to \infty} [\arctan (R)-\arctan (0)]\\=0+\dfrac{\pi}{2}+\dfrac{\pi}{2}-0 \\= \pi$