Evaluate $\lim _{n\to \infty }n\int _1^2 \frac{dx}{x^2(1+x^n)}$ Evaluate $$\lim _{n\to \infty }n\int _1^2 \frac{dx}{x^2(1+x^n)}$$
without Taylor expansion.
I tried rewriting as 
$$\lim_{n\to \infty} n\int _1^2 \frac{1 + x^n - x^n}{x^2(1+x^n)}dx = \lim_{n\to \infty} n\int _1^2 \frac{dx}{x^2} - \lim_{n\to \infty} n\int _1^2 \frac{x^n}{x^2(1+x^n)}dx$$
The first integral is computable, but I don't know how to continue solving the second one.
The answer is $\ln 2$
 A: By setting $x=z^{1/n}$ we are left with
$$ \lim_{n\to +\infty}\int_{1}^{2^n}\frac{dz}{z(1+z)z^{1/n}}=\lim_{n\to +\infty}\left[O\left(\frac{1}{2^n}\right)+\int_{1}^{+\infty}\left(\frac{1}{z}-\frac{1}{z+1}\right)\frac{dz}{z^{1/n}}\right] $$
then by applying the dominated convergence theorem we get that the answer is $\color{red}{\log 2}$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\lim_{n \to \infty}\bracks{n\int_{1}^{2}{\dd x \over x^{2}\pars{1 + x^{n}}}} =
\lim_{n \to \infty}\bracks{n\int_{1/2}^{1}{x^{n} \over 1 + x^{n}}\,\dd x} =
\lim_{n \to \infty}\bracks{\int_{1/2^{n}}^{1}{x^{1/n} \over 1 + x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{\int_{0}^{1}{x^{1/n} \over 1 + x}\,\dd x -
\int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{\int_{0}^{1}{\dd x \over 1 + x} -
\int_{0}^{1}{1 - x^{1/n} \over 1 + x}\,\dd x -
\int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x}
\\[5mm] = &\
\lim_{n \to \infty}\bracks{\ln\pars{2} - H_{1/n} - 
\int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x}\qquad
\pars{~H_{z}:\ Harmonic\ Number.\ \mbox{Note that}\ H_{0} = 0~}
\end{align}

Moreover,
  $\ds{0 < \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x < 
\int_{0}^{1/2^{n}}{x^{1/n} \over 1 + 0}\,\dd x =
{n \over n + 1}\,{1 \over 2^{n + 1}}\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0}}$


$$
\bbx{\ds{\lim_{n \to \infty}
\bracks{n\int_{1}^{2}{\dd x \over x^{2}\pars{1 + x^{n}}}} =
\ln\pars{2}}}
$$
