Integration of 1/(1+x^2)
Formulas
- Definite Integral
$$ \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx = \pi $$
- Indefinite Integral
$$ \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx = \pi $$
$C$ is the integration constant.
Proofs
Definite Integral
Let’s substitute with $x = \tan \theta$. Then, the range of integration becomes $\displaystyle \int_{-\infty}^{\infty} \to \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$, and since $\tan ^{\prime} = \sec^{2}$, it results in $dx = \sec^{2} d\theta$.
$$ \begin{align*} \int_{-\infty}^{\infty} \dfrac{1}{1+x^{2}}dx &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{1 + \frac{\sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{\cos^{2} \theta + \sin^{2} \theta}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \dfrac{1}{\frac{1}{\cos^{2} \theta}} \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \sec^{2} \theta d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} d\theta \\ &= \frac{\pi}{2} - (-\frac{\pi}{2}) \\ &= \pi \end{align*} $$
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Indefinite Integral
Similarly, by substituting with $x = \tan \theta$,
$$ \begin{align*} \int \dfrac{1}{1+x^{2}}dx &= \int \dfrac{1}{1 + \tan^{2}\theta} \sec^{2} \theta d\theta \\ &= \int \cos^{2} \theta \dfrac{1}{\cos^{2} \theta} d\theta \\ &= \int d\theta \\ &= \theta + C \\ &= \tan^{-1} x + C \end{align*} $$
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