Explanation

(d)
Let $I = \int_{e^{- 1}}^{e}\frac{dt}{t(t + 1)}$
$= \int_{e^{- 1}}^{e}{\frac{1}{t}dt - \int_{e^{- 1}}^{e}{\frac{1}{(t + 1)}dt}}$
$= \left\lbrack \log{t - \log(t + 1)} \right\rbrack_{e^{- 1}}^{e}$
$= \lbrack\log{e - log(e + 1)\rbrack - \lbrack\log{e^{- 1} - log(e^{- 1} + 1)\rbrack}}$
$= \log\left( \frac{e}{e + 1} \right) - \lbrack - \log{e - \log(e^{- 1} + 1)}\rbrack$
$= \log{\left( \frac{e}{e + 1} \right) + \left\lbrack \log e\left( \frac{1}{e} + 1 \right) \right\rbrack}$
$= \log_{e}{e = 1}$