Question

integrala din( x/(x^2+x+1) )

Answer 1

$\displaystyle \int \dfrac{x}{x^2+x+1}\, dx=$

$\displaystyle = \dfrac{1}{2}\int \dfrac{2x}{x^2+x+1}\, dx$

$\displaystyle= \dfrac{1}{2}\int \dfrac{2x+1-1}{x^2+x+1}\, dx$

$\displaystyle = \dfrac{1}{2}\int \left(\dfrac{2x+1}{x^2+x+1}-\dfrac{1}{x^2+x+1}\right)\, dx$

$\displaystyle = \dfrac{1}{2}\int \dfrac{2x+1}{x^2+x+1}\, dx - \dfrac{1}{2}\int \dfrac{1}{x^2+x+1}\, dx$

$\displaystyle = \dfrac{1}{2}\int \dfrac{(x^2+x+1)'}{x^2+x+1}\, dx-\dfrac{1}{2}\int \dfrac{1}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}\, dx$

$\displaystyle = \dfrac{1}{2}\ln(x^2+x+1)-\dfrac{1}{2}\int \dfrac{1}{\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\, dx$

$\displaystyle =\dfrac{1}{2}\ln(x^2+x+1)-\dfrac{1}{2}\cdot \dfrac{2}{\sqrt{3}}\,\text{arctg}\left(\dfrac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$

$\displaystyle=\dfrac{1}{2}\ln(x^2+x+1)-\dfrac{1}{\sqrt{3}}\,\text{arctg}\left(\dfrac{2x+1}{\sqrt{3}}\right)+C$