Risolvere senza la tecnica X SOSTITUZIONE.
Gentilmente spiegare i passaggi.
$\int_2^{2\sqrt{3}} \frac{k+1}{x^2+4} \, dx = \frac{\pi}{6} $
$ \frac{k+1}{4} \int_2^{2\sqrt{3}} \frac{1}{1+(\frac{x}{2})^2} \, dx = \frac{\pi}{6} $
Rendiamolo immediato
$ \frac{k+1}{2} \int_2^{2\sqrt{3}} \frac{1}{1+(\frac{x}{2})^2} \frac{1}{2} \, dx = \frac{\pi}{6} $
$ \frac{k+1}{2} \left. arctan(\frac{x}{2}) \right|_2^{2\sqrt{3}} = \frac{\pi}{6}$
$ \frac{k+1}{2} [\frac{\pi}{3} - \frac{\pi}{4}] = \frac{\pi}{6}$
$ \frac{k+1}{2} \frac{\pi}{12} = \frac{\pi}{6}$
$ \frac{k+1}{2} = 2$
$ k = 3$