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8.1: La función Arcangente
https://espanol.libretexts.org/Matematicas/Analisis/Libro%3A_Una_cartilla_de_an%C3%A1lisis_real_(Sloughter)/08%3A_M%C3%A1s_funciones/8.01%3A_La_funci%C3%B3n_Arcangente
Usando la sustitución $t=\frac{1}{u},$ que tenemos \[\begin{aligned} \arctan \left(\frac{1}{x}\right) &=\int_{0}^{\frac{1}{x}} \frac{1}{1+t^{2}} d t \\ &=\int_{+\infty}^{x} \frac{1}{1+\frac{1}{u^{2}}...Usando la sustitución $t=\frac{1}{u},$ que tenemos $\begin{aligned} \arctan \left(\frac{1}{x}\right) &=\int_{0}^{\frac{1}{x}} \frac{1}{1+t^{2}} d t \\ &=\int_{+\infty}^{x} \frac{1}{1+\frac{1}{u^{2}}}\left(-\frac{1}{u^{2}}\right) d u \\ &=-\int_{+\infty}^{x} \frac{1}{1+u^{2}} d u \\ &=\int_{x}^{+\infty} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\int_{0}^{x} \frac{1}{1+u^{2}} d u \\ &=\frac{\pi}{2}-\arctan (x). \end{aligned}$ Q.E.D.

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