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




Salut Ma Puteti Ajuta Unde GresescAsa Am Invatat La Scoala Si Asa Fac Si Alte Probleme Insa Asta Nu Iasa Ma Puteti Ajuta class=

Răspuns :

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)!

[tex]\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 =[/tex]

[tex]=0-\dfrac{1}{\sqrt 2}\cdot\Big(- \dfrac{\pi}{2}\Big) = \dfrac{\pi}{2\sqrt 2}[/tex]