Question

Salut, aveti o idee la problema 717...?

Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle I= \int_{0}^1\dfrac{1+x^2}{1+x^2+x^4}\, dx = \int_{0}^1\dfrac{x^{-2}}{x^{-2}}\cdot \dfrac{1+x^2}{1+x^2+x^4}\, dx = \\ \\ =\int_{0}^1 \dfrac{1+x^{-2}}{x^{2}+x^{-2}+1}\, dx= \int_{0}^1 \dfrac{1+x^{-2}}{(x-x^{-1})^2+3}\, dx\\ \\\\ x-x^{-1} = t \Rightarrow (1+x^{-2})\, dx = dt\\ x=0 \Rightarrow t\to -\infty \\ x = 1 \Rightarrow t =0 \\ \\ I =\int_{-\infty}^{0}\dfrac{1}{t^2+3}\, dt = \dfrac{1}{\sqrt 3}\,\mathrm{arctg}\Big(\dfrac{t}{3}\Big)\Big|_{-\infty}^0=$

$=0 - \dfrac{1}{\sqrt 3}\cdot \Big(-\dfrac{\pi}{2}\Big)$

$\Rightarrow \boxed{I =\dfrac{\pi}{2\sqrt 3}}$