Question

Salut, ma puteti ajuta unde gresesc...Asa am invatat la scoala si asa fac si alte probleme insa asta nu iasa, ma puteti ajuta...?

Answer 1

tg pi/4=1, NU invers!

integrala se rezolva prin metoda fractii simple, numitorul de jos se scrie ca (x^2-radical(2)*x+1)(x^2+radical(2)*x+1)!

Answer 2

$\displaystyle \int_0^1 \dfrac{x^2+1}{x^4+1}\, dx = \int_0^1 \dfrac{x^{-2}}{x^{-2}}\cdot \dfrac{x^2+1}{x^4+1}\, dx = \int_0^1 \dfrac{1+x^{-2}}{x^2+x^{-2}-2+2}\, dx = \\ \\ = \int_0^1\dfrac{1+x^{-2}}{(x-x^{-1})^2+2}\, dx= \\ \\ \\x-x^{-1} = u \Rightarrow (1+x^{-2})dx = du \\ x = 1 \Rightarrow u\to 0 \\ x = 0 \Rightarrow u \to -\infty \\ \\ = \int_{-\infty}^0 \dfrac{1}{u^2+(\sqrt{2})^2}\, du = \dfrac{1}{\sqrt 2}\arctan\Big(\dfrac{u}{\sqrt 2}\Big)\Big|_{-\infty}^0 =$

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