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\begin{align}
\int_{0}^{1}{x^{n - 1} \over 1 + x}\,\dd x&
=\sum_{k\ =\ 0}^{\infty}\pars{-1}^{k}\int_{0}^{1}x^{n - 1 + k}\,\dd x
=\sum_{k\ =\ 0}^{\infty}{\pars{-1}^{k} \over n + k}
=\sum_{k\ =\ 0}^{\infty}\pars{{1 \over 2k + n} - {1 \over 2k + 1 + n}}
\\[5mm]&={1 \over 4}\sum_{k\ =\ 0}^{\infty}
{1 \over \bracks{k + \pars{n + 1}/2}\pars{k + n/2}}
=\half\bracks{\Psi\pars{n + 1 \over 2} - \Psi\pars{n \over 2}}
\end{align}
where $\ds{\Psi}$ is the Digamma Function.
Then,
\begin{align}
\color{#66f}{\large a}&=\lim_{n\ \to\ \infty}
n\braces{\half\bracks{\Psi\pars{n + 1 \over 2} - \Psi\pars{n \over 2}}}
=\color{#66f}{\large\half}
\\[5mm]
\color{#66f}{\large b}&=\lim_{n\ \to\ \infty}
n^{2}\braces{
\half\bracks{\Psi\pars{n + 1 \over 2} - \Psi\pars{n \over 2}} - {1 \over 2n}}
=\color{#66f}{\large{1 \over 4}}
\end{align}
