Explanation

(a)
Let $I = \int_{}^{}{\frac{\log{(x + 1) - \log x}}{x(x + 1)}dx} = \int_{}^{}{\frac{\log\left( 1 + \frac{1}{x} \right)}{x(x + 1)}dx}$
$= \int_{}^{}\frac{\log\left( 1 + \frac{1}{x} \right)}{x^{2}\left( 1 + \frac{1}{x} \right)}dx$
Put $1 + \frac{1}{x} = t\ \ \ \Rightarrow \ \ - \frac{1}{x^{2}}\ dx = dt$
$\therefore\ I = - \int_{}^{}{\frac{\log t}{t}dt} = - \frac{1}{2}\left( \log t \right)^{2} + c$
$= - \frac{1}{2}\left\lbrack \log\left( 1 + \frac{1}{x} \right) \right\rbrack^{2} + c$
$= - \frac{1}{2}\left\lbrack \log{(x + 1) - \log x} \right\rbrack^{2} + c$
$= - \frac{1}{2}\left\{ \log(x + 1) \right\}^{2} - \frac{1}{2}\left( \log x \right)^{2} + \log(x + 1).\log x + c$