[tex] \frac{x^2}{x^4+x^2+1} = \frac{x^2}{(x^2+1)^2-x^2}= \frac{x^2}{(x^2-x+1)(x^2+x+1)} = \frac{Ax+B}{x^2-x+1} + \frac{Cx+D}{x^2+x+1} [/tex]
Daca identificam coeficientii si rezolvam sistemul de ecuatii obtinem:A=1/2,B=0,C=-1/2,D=0.
[tex]I= \frac{1}{2} \int\limits^3_0 {\frac{x}{x^2-x+1}} \, dx - \frac{1}{2} \int\limits^3_0 {\frac{x}{x^2+x+1}} \, dx \\
I= \frac{1}{4} \int\limits^3_0 {\frac{2x-1+1}{x^2-x+1}} \, dx - \frac{1}{4} \int\limits^3_0 {\frac{2x+1-1}{x^2+x+1}} \, dx \\
I= \frac{1}{4} ( \int\limits^3_0 { \frac{2x-1}{x^2-x+1} } \, dx + \int\limits^3_0 { \frac{1}{x^2-x+1} } \, dx - \int\limits^3_0 { \frac{2x+1}{x^2+x+1} } \, dx+\int\limits^3_0 { \frac{1}{x^2+x+1} } \, dx)\\
[/tex]
[tex]I= \frac{1}{4} ( \int\limits^7_1 { \frac{1}{t} } \, dt+ \int\limits^3_0 { \frac{1}{(x- \frac{1}{2})^2+( \frac{\sqrt{3}}{2} )^2 } } \, dx - \int\limits^{13}_1 {\frac{1}{t} \, dt + \int\limits^3_0 {\frac{1}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}} \, dx )[/tex]
In final obtinem:
[tex]I=\frac{1}{4}[ln7+\frac{2}{\sqrt{3}}(arctg\frac{5}{\sqrt{3}}+arctg\frac{1}{\sqrt{3}})-ln13+\frac{2}{\sqrt{3}}(arctg\frac{7}{\sqrt{3}}-arctg\frac{1}{\sqrt3}}]}[/tex]
[tex]I=\frac{1}{4}[ln\frac{7}{13}+\frac{2}{\sqrt{3}}(arctg\frac{5}{\sqrt{3}}+arctg\frac{7}{\sqrt{3}})][/tex]
[tex]I\approx0,585852[/tex]
