Question

Salut, ma puteti ajuta la problema 718, am incercat prin parti insa nu iasa...

Answer 1

Răspuns:

Explicație pas cu pas:

Rezolvarea integrală este în imaginea de mai jos. Mult succes!

Answer 2

$I = \displaystyle \int_{\frac{1}{n}}^n \dfrac{\text{arctg}\,(x^2)}{1+x^2}\, dx \\ \\ x = \dfrac{1}{t} \Rightarrow dx = -\dfrac{1}{t^2}\,dt \\ \\ x = \dfrac{1}{n}\Rightarrow t = n\\ x = n \Rightarrow t = \dfrac{1}{n}\\ \\ I = \int_{n}^{\frac{1}{n}}\dfrac{\text{arctg} (\frac{1}{t^2})}{1+\frac{1}{t^2}}\cdot (-\dfrac{1}{t^2})\, dt = \int_{\frac{1}{n}}^n \dfrac{t^2\text{arctg}(\frac{1}{t^2})}{1+t^2}\cdot \dfrac{1}{t^2}\, dt =$

$\displaystyle = \int_{\frac{1}{n}}^n\dfrac{\text{arctg}(\frac{1}{t^2})}{1+t^2}\, dt \\ \\ I + I = \int_{\frac{1}{n}}^n \dfrac{\text{arctg}\,(x^2)}{1+x^2}\, dx + \int_{\frac{1}{n}}^n \dfrac{\text{arctg}\,(\frac{1}{x^2})}{1+x^2}\, dx \\ \\ 2I = \int_{\frac{1}{n}}^n \dfrac{\text{arctg}\,(x^2)+\text{arctg}(\frac{1}{x^2})}{1+x^2}\, dx\\ \\ 2I = \int_{\frac{1}{n}}^n\dfrac{\dfrac{\pi}{2}}{1+x^2}\, dx$

$2I = \dfrac{\pi}{2}\int_{\frac{1}{n}}^n\dfrac{1}{1+x^2}\, dx\\ \\ 2I = \dfrac{\pi}{2}\cdot \text{arctg}(x)\Big|_{\frac{1}{n}}^n \\ \\ I = \dfrac{\pi}{4}\cdot \text{arctg}(n) - \dfrac{\pi}{4}\cdot \text{arctg}(\frac{1}{n})\\ \\ \Rightarrow \lim\limits_{n\to \infty }I = \dfrac{\pi}{4}\cdot \dfrac{\pi}{2}-0 = \boxed{\dfrac{\pi^2}{8}}$