[tex]f(x)=1+\frac{1}{\sqrt{x^{2}+1}}[/tex]
a)
Vezi tabelul de integrale din atasament
[tex]\int\limits^1_0 {1+\frac{1}{\sqrt{x^2+1} } } \, dx =x|_0^1+ln(x+\sqrt{x^2+1})|_0^1 =1+ln(1+\sqrt{2})-ln1=1+ ln(1+\sqrt{2})[/tex]
b)
[tex]\int\limits^1_{-1} {|x+\frac{x}{\sqrt{x^2+1} } |} \, dx =\int\limits^0_{-1} {-x} \, dx +\int\limits^1_0 {x} \, dx+ \int\limits^0_{-1} \frac{-x}{\sqrt{x^2+1} } \ dx+ \int\limits^1_{0} \frac{x}{\sqrt{x^2+1} } \ dx=-\frac{x^2}{2}|_{-1}^0 +\frac{x^2}{2}|_0^1- \sqrt{x^2+1}|_{-1}^0+\sqrt{x^2+1}|_0^1=\\\\ =0+\frac{1}{2}+\frac{1}{2} -(1-\sqrt{2} )+\sqrt{2} -1=\\\\ 1-1+2\sqrt{2}-1=2\sqrt{2} -1[/tex]
c)
[tex]\lim_{x\to 0} \frac{1}{x} \int\limits^x_0 {1+\frac{1}{\sqrt{t^2+1} } } \, dx[/tex]
Aplicam l'Hopital, derivam numarator, derivam numitor si obtinem:
[tex]\lim_{x\to 0} \frac{f(x)}{1}=f(0)=1+1=2[/tex]
Un alt exercitiu cu integrale gasesti aici: https://brainly.ro/tema/9905358
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