[tex]f(x)=\frac{x^{2}+4}{x}[/tex]
a)
[tex]\int\limits^3_1 {\frac{x^2+4}{x}-\frac{4}{x} } \, dx =\int\limits^3_1 \frac{x^2}{x} \ dx=\int\limits^3_1x\ dx=\frac{x^2}{2} |_1^3=\frac{9}{2}-\frac{1}{2} =4[/tex]
b)
[tex]\int\limits^6_2{\frac{2}{\frac{x^2+4}{x} } } \, dx =\int\limits^6_2{\frac{2x}{x^2+4} } } \, dx[/tex]
Daca partea de jos a fractiei (numitorul) derivata este egala cu numaratorul atunci intregrala noastra este egala cu ln(numitor) - vezi tabel integrale compuse din atasament
(x²+4)'=2x
[tex]\int\limits^6_2{\frac{2}{\frac{x^2+4}{x} } } \, dx =\int\limits^6_2{\frac{2x}{x^2+4} } } \, dx=ln(x^2+4)|_2^6=ln40-ln8=ln5[/tex]
c)
[tex]\int\limits^e_1 {(\frac{x^2+4}{x}-\frac{4}{x} )lnx} \, dx =\int\limits^e_1 xlnx\ dx[/tex]
Facem prin integrarea prin parti
[tex]f=lnx\ \ \ \ \ \ \ f'=\frac{1}{x} \\\\g'=x\ \ \ \ \ \ \ g=\frac{x^2}{2}[/tex]
[tex]\int\limits^e_1 xlnx\ dx=\frac{x^2}{2}lnx|_1^e- \int\limits^e_1 \frac{x^2}{2x} \ dx=\frac{x^2}{2}lnx|_1^e- \int\limits^e_1\frac{x}{2} \ dx=\frac{x^2}{2}lnx|_1^e- \frac{x^2}{4}|_1^e=\\\\=\frac{e^2}{2}lne-0-\frac{e^2}{4}+\frac{1}{4}= \frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}\\\\ \frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}=\frac{e^2+1}{a} \\\\\frac{e^2+1}{4}=\frac{e^2+1}{a}\\\\[/tex]
a=4
Un alt exercitiu similar de bac il gasesti aici: https://brainly.ro/tema/956588
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