Explanation

(b)
Let $I = \int_{1}^{2}\frac{dx}{x(1 + x^{4})} = \int_{1}^{2}{\frac{x^{3}}{x^{4}(1 + x^{4})}dx}$
Put $x^{4} = t \Rightarrow 4x^{3}dx = dt$
$\therefore\ \ \ \ I = \frac{1}{4}\int_{1}^{16}{\frac{dt}{t(1 + t)} = \frac{1}{4}\int_{1}^{16}{\left( \frac{1}{t} - \frac{1}{1 + t} \right)dt}}$
$= \frac{1}{4}\left( \log\frac{t}{1 + t} \right)_{1}^{16} = \frac{1}{4}\left( \log{\frac{16}{17} - \log\frac{1}{2}} \right)$
$= \frac{1}{4}\log\frac{32}{17}$