Question

$\int\limits^3_0 { \frac{ x^{2} }{ x^{4}+ x^{2} +1} } \, dx$

Answer 1

f= x^4+x^2+1 ; f'=6x
g=x^2; g'=2x
S-integrala nu prea stiu sa folosesc semnele pe site, sunt nou.

S de la 0 la 3 (f ' * g - f * g ')/g^2 dx .
Dupa calcule si simplificari vei ajunge la
S de la 0 la 3 (2x^5 +4x^3 -2x)/x^4 dx . Cand ajungi la forma asta imparti ce e sus la ce e jos ca pe un polinom . Dupa pui rezultatul in integrala si faci o integrala simpla :) . Nu prea am explicat bine dar nu sunt profesor si din pacate nu stiu sa folosesc semnele ai fi inteles mai bine daca iti scriam pe foaie.

Answer 2

$\frac{x^2}{x^4+x^2+1} = \frac{x^2}{(x^2+1)^2-x^2}= \frac{x^2}{(x^2-x+1)(x^2+x+1)} = \frac{Ax+B}{x^2-x+1} + \frac{Cx+D}{x^2+x+1}$
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]
$I= \frac{1}{4} ( \int\limits^7_1 { \frac{1}{t} } \, dt+ \int\limits^3_0 { \frac{1}{(x- \frac{1}{2})^2+( \frac{\sqrt{3}}{2} )^2 } } \, dx - \int\limits^{13}_1 {\frac{1}{t} \, dt + \int\limits^3_0 {\frac{1}{(x+\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}} \, dx )$
In final obtinem:
$I=\frac{1}{4}[ln7+\frac{2}{\sqrt{3}}(arctg\frac{5}{\sqrt{3}}+arctg\frac{1}{\sqrt{3}})-ln13+\frac{2}{\sqrt{3}}(arctg\frac{7}{\sqrt{3}}-arctg\frac{1}{\sqrt3}}]}$
$I=\frac{1}{4}[ln\frac{7}{13}+\frac{2}{\sqrt{3}}(arctg\frac{5}{\sqrt{3}}+arctg\frac{7}{\sqrt{3}})]$
$I\approx0,585852$