Question

Salut, aveti o idee de rezolvare la problema 690...Am tot incercat insa nu-mi iasa nimic

Answer 1

$\text{Se foloseste urmatoarea descompunere in factori:}\\ x^4+x^2+1=x^2+2x^2+1-x^2=(x^2+1)^2-x^2= \\ =\boxed{(x^2+x+1)(x^2-x+1)}\\ \text{Prin urmare:}\\ \displaystyle\int \dfrac{x^2+1}{x^2+x^2+1}dx=\displastyle\int \left(\dfrac{1}{2(x^4+x+1)}-\dfrac{1}{2(x^2-x+1)}\right)dx= \\ =\dfrac{1}{2}\left(\displatyle \int \dfrac{1}{x^2+x+1}dx-\int \dfrac{1}{x^2-x+1}dx\right)=\\ \text{Mai ramane sa rezolvam cele doua integrale(se rezolva asemanator):}$

$\displaystyle\int \dfrac{1}{x^2+x+1}dx= \int \dfrac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}dx \\ \text{Substituim:}\\ u=\dfrac{2x+1}{\sqrt 3}\Rightarrow dx=\dfrac{2}{\sqrt 3}du\\ \text{Integrala va deveni:}\\ \int \dfrac{2\sqrt 3}{3u^2+3} du= \dfrac{2}{\sqrt 3}\int \dfrac{1}{u^2+1}du =\dfrac{2}{\sqrt 3} \cdot \arctan u +C =\\ \dfrac{2\cdot \arctan\left(\frac{2x+1}{\sqrt 3}\right)}{\sqrt 3}+C\\ \text{Pentru cea de-a doua procedand analog se obtine: }$

$\int \dfrac{1}{x^2-x+1}dx=\dfrac{2\cdot \arctan \left(\frac{2x-1}{\sqrt 3} \right)}{\sqrt 3}+C\\ \text{Asadar:}\\ \dfrac{1}{2} \left(\displaystyle \int \dfrac{1}{x^2+x+1}dx-\int \dfrac{1}{x^2-x+1}dx\right)= \\ = \boxed{ \dfrac{\arctan\left(\frac{2x+1}{3}\right)+\arctan \left(\frac{2x-1}{\sqrt 3}\right)}{\sqrt 3}+C}\\ \text{Dupa ce faci faci calculele iti va da } \dfrac{\pi }{2\sqrt 3} \text{deci raspunsul este B.}$