[tex]f(x)=\frac{x}{\sqrt{x^{2}+1}}[/tex]
a)
Vezi tabelul de integrale din atasament
[tex]\int\limits^2_0 {x} \, dx =\frac{x^2}{2}|_0^2=\frac{4}{2} =2[/tex]
b)
[tex]\int\limits^2_1 {\frac{x}{\sqrt{x^{2}+1}}+\frac{\frac{1}{x} }{\sqrt{\frac{1}{x^2}+1 } } } \, dx =\int\limits^2_1 {\frac{x}{\sqrt{x^{2}+1}}\ dx+\int\limits^2_1 {\frac{\frac{1}{x} }{\sqrt{\frac{1+x^2}{x^2} } } }\ dx=\sqrt{x^2+1} |_1^2+\int\limits^2_1 {\frac{\frac{1}{x} }{\sqrt{\frac{1+x^2}{x^2} } } }\ dx[/tex]
Luam integrala separat
[tex]\int\limits^2_1 {\frac{\frac{1}{x} }{\sqrt{\frac{1+x^2}{x^2} } } }\ dx=\int\limits^2_1\frac{1}{\sqrt{x^2+1} } \ dx=ln(x+\sqrt{x^2+1})|_1^2=ln(2+\sqrt{5} )-ln(1+\sqrt{2})[/tex]
Deci integrala noastra din cerinta va fi egala cu:
[tex]\sqrt{x^2+1} |_1^2+ln(2+\sqrt{5} )-ln(1+\sqrt{2}) =\sqrt{5} -\sqrt{2}+ln\frac{2+\sqrt{5} }{1+\sqrt{2} }[/tex]
c)
[tex]\int\limits^x_0 {\frac{e^t}{\sqrt{e^{2t}}+1 } } \, dt=ln(e^t+\sqrt{e^{2t}+1} )|_0^x=ln(e^x+\sqrt{e^{2x}+1}) -ln(1+\sqrt{2}) \\\\ln(e^x+\sqrt{e^{2x}+1}) -ln(1+\sqrt{2}) =ln(e^x+\sqrt{e^{2x}+1}) +ln(a-1)\\\\-ln(1+\sqrt{2})=ln(a-1)\\\\a-1=\frac{1}{1+\sqrt{2} } \\\\a=\frac{1}{1+\sqrt{2} } +1=\frac{1-\sqrt{2} }{-1}+1=\sqrt{2}[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9919133
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