[tex]f(x)=\frac{x^{3}-x-1}{x^{2}(x+1)}[/tex]
[tex]F(x)=\frac{x^{2}+1}{x}-\ln (x+1)[/tex]
a)
F este primitiva a functiei f
F'(x)=f(x)
Vezi tabelul de derivate si integrale in atasament
[tex]F'(x)=\frac{2x\cdot x-(x^2+1)}{x^2} -\frac{1}{x+1} =\frac{(2x^2-x^2-1)(x+1)-x^2}{x^2(x+1)} =\frac{x^3+x^2-x-1-x^2}{x^2(x+1)}=\frac{x^3-x-1}{x^2(x+1)} =f(x)[/tex]
b)
[tex]\int\limits^2_1 {\frac{x^3-x-1}{x^2} } \, dx[/tex]
"Spargem" integrala in 3 integrale si obtinem
[tex]\int\limits^2_1 x\ dx-\int\limits^2_1 \frac{1}{x}\ dx-\int\limits^2_1 \frac{1}{x^2}\ dx=\frac{x^2}{2}|_1^2-lnx|_1^2-\frac{x^{-1}}{-1} |_1^2=2-\frac{1}{2}-ln2+\frac{1}{2} -1=1-ln2[/tex]
c)
Ne folosim de punctul a
[tex]\int\limits^a_1 {f(x)} \, dx =F(x)|_1^a=F(a)-F(1)\\\\\frac{a^2+1}{a}-ln(a+1)-2+ln2=\frac{1}{2}-ln\frac{a+1}{2} \\\\ \frac{a^2+1}{a}-2-ln\frac{a+1}{2}=\frac{1}{2}-ln\frac{a+1}{2}\\\\ \frac{a^2+1}{a}-2=\frac{1}{2}\\\\[/tex]
Aducem la acelasi numitor comun 2a
2a²+2-4a=a
2a²-5a+2=0
Δ=25-16=9
[tex]a_1=\frac{5+3}{4} =2\\\\a_2=\frac{5-3}{4} =\frac{1}{2}\ NU[/tex]
a=2
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9918943
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