[tex]\boxed{Acesta \:\ este \:\ punctul \:\ C}[/tex]
[tex]\int^n_0 \frac{dx}{(x+1)(x+2)} = \int \frac{1}{\left(x+1\right)\left(x+2\right)}dx
=\int \frac{1}{x+1}-\frac{1}{x+2}dx =\int \frac{1}{x+1}dx-\int \frac{1}{x+2}dx
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\\Schimbare \:\ de \:\ variabila: =\ \textgreater \ \ln \left|x+1\right|-\ln \left|x+2\right|
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\\=\ \textgreater \ \frac{1}{2}\ln \left(\frac{\left(n+1\right)^2}{\left(n+2\right)^2}\right)-\left(-\ln \left(2\right)\right) =\ \textgreater \ \boxed{simplificare} = \ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)
[/tex]
[tex]\\ =\ \textgreater \ \lim_{\rightarrow\infty} n[ln2-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)]
\\ =\ \textgreater \ \lim _{n\to a}\left[f\left(n\right)\cdot g\left(n\right)\right]=\lim _{n\to a}f\left(n\right)\cdot \lim _{n\to a}g\left(n\right)
\\=\ \textgreater \ \lim _{n\to \infty \:}\left(n\right)\cdot \lim \:_{n\to \infty \:}\left(\ln \left(2\right)-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)\right)
\\=\ \textgreater \ \lim _{n\to \infty \:}\left(n\right)=\infty \:
[/tex]
[tex]\\ si \:\ \lim _{n\to \infty \:}\left(\ln \left(2\right)-\ln \left(\frac{n+1}{n+2}\right)+\ln \left(2\right)\right)
\\= \lim _{n\to \infty \:}\left(\ln \left(2\right)\right)-\lim _{n\to \infty \:}\left(\ln \left(\frac{n+1}{n+2}\right)\right)+\lim _{n\to \infty \:}\left(\ln \left(2\right)\right)
\\ \lim _{n\to \infty \:}\left(\ln \left(\frac{n+1}{n+2}\right)\right) =\ \textgreater \ g\left(n\right)=\frac{n+1}{n+2},\:f\left(u\right)=\ln \left(u\right)
[/tex]
[tex]\lim _{n\to \infty \:}\left(\frac{n+1}{n+2}\right) =\lim _{n\to \infty \:}\left(\frac{1+\frac{1}{n}}{1+\frac{2}{n}}\right) =\frac{\lim _{n\to \infty \:}\left(1+\frac{1}{n}\right)}{\lim _{n\to \infty \:}\left(1+\frac{2}{n}\right)} = \frac{1}{1} = 1
\\ \lim _{u\to \:1}\left(\ln \left(u\right)\right) =\ln \left(1\right) = 0
\\ Deci \:\ rezultatul \:\ final \:\ este \:\ =\ln \left(2\right)-0+\ln \left(2\right)
\\ =2\ln \left(2\right) =\infty \cdot \:2\ln \left(2\right) = \boxed{\infty \: }[/tex]
