Explanation

(a)
$F = (e) = f(e) + f\left( \frac{1}{e} \right)$
$\Rightarrow F(e) = \int_{1}^{e}\frac{\log t}{1 + t}dt + \int_{1}^{1/e}{\frac{\log t}{1 + t}dt}$
On putting $t = \frac{1}{t}$ in second integration, we get
$F(e) = \int_{1}^{e}\frac{\log t}{1 + t}dt + \int_{1}^{e}{\frac{\log t}{t(1 + t)}dt}$
$= \int_{1}^{e}\frac{\log t}{t}dt = \left\lbrack \frac{(\log{t\left. \ \ \right)^{2}}}{2} \right\rbrack_{1}^{e}$
$= \frac{1}{2}\lbrack(\log{e\left. \ \ \right)^{2} - (\log{1\left. \ \ \right)^{2}\rbrack = \frac{1}{2}}}$