The integral \(\Large\int\limits_{1}^{\infty}\frac{dx}{x(x+1)}\)
Correct Answer: Description for Correct answer:
\(\Large \int\limits_{1}^{\infty}\frac{dx}{x(x+1)}=\lim\limits_{t\rightarrow \infty}\int\limits_{1}^{t}\frac{dx}{x(x+1)}\)
\(\Large =\lim\limits_{t\rightarrow \infty}\int\limits_{1}^{t} \left(\frac{1}{x}-\frac{1}{x+1}\right)dx \)
\(\Large=\lim\limits_{t\rightarrow \infty}\left[ log x-log(1+x) \right]_{1}^{t}\)
\(\Large =\lim\limits_{t\rightarrow \infty}\left[ log\frac{x}{x+1} \right]_{1}^{t}\)
\(\Large =\lim\limits_{t\rightarrow \infty}\left[ log \left(\frac{1}{1+\frac{1}{t}}\right)-log\frac{1}{2}\right]\)
\(\Large =log1-log\frac{1}{2}=-log\frac{1}{2}\)
\(\Large =log2\)
Therefore the integral converges to \(log2.\)
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