[tex]f(x)=4 x-\frac{2 x}{x^{2}+1}+\frac{1}{x^{2}+1}[/tex]
a)
[tex]\int\limits^1_0 {(x^2+1)f(x)} \, dx =\int\limits^1_04x(x^2+1)-2x+1\ dx=\int\limits^1_04x^3+2x+1\ dx=\frac{4x^4}{4} |_0^1+x^2|_0^1+x|_0^1=1+1+1=3[/tex]
b)
[tex]\int\limits^1_0 {4 x-\frac{2 x}{x^{2}+1}+\frac{1}{x^{2}+1}} \, dx =2x^2|_0^1-ln(x^2+1)|_0^1+arctgx|_0^1=2-ln2+\frac{\pi}{4}[/tex]
(Vezi in atasament formulele de integrare si derivare)
c)
[tex]\int\limits^e_1 {(4x-\frac{2x}{x^2+1} +\frac{1}{x^2+1}+\frac{2x-1}{x^2+1})lnx } \, dx=\int\limits^e_1 4x lnx\ dx=4\int\limits^e_1 x lnx\ dx[/tex]
Notam [tex]I=\int\limits^e_1 x lnx\ dx[/tex]
O integram prin parti
[tex]f=lnx\ \ \ \ f'=\frac{1}{x} \\\\g'=x\ \ \ \ g=\frac{x^2}{2}[/tex]
[tex]I=\frac{x^2}{2}lnx|_1^e -\frac{1}{2}\int\limits^e_1 {x} \, dx \\\\I=\frac{x^2}{2}lnx|_1^e -\frac{1}{2}\cdot \frac{x^2}{2}|_1^e \\\\I=\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4}[/tex]
[tex]4I=4(\frac{e^2}{2}-\frac{e^2}{4}+\frac{1}{4} )\\\\2e^2-e^2+1=e^2+a\\\\a=1[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9882193
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