Explanation:
(C) : Given,
\(f(x)=\int_{1}^{x} \frac{\log _{e} t}{(1+t)} d t\)
\(f(e)=\int_{1}^{e} \frac{\log _{e} t}{1+t} d t\)
\(f\left(\frac{1}{e}\right)=\int_{1}^{1 / e} \frac{\log _{e} t}{1+t} d t \tag{ii}\)
Let, \(\quad t=\frac{1}{u}\) and putting in equation (ii), we get-
\(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\int_{1}^{\mathrm{e}} \frac{\log \left(\frac{1}{\mathrm{u}}\right)}{1+\frac{1}{\mathrm{u}}} \times \frac{-1}{\mathrm{u}^{2}} \mathrm{du}\)
\(=\int_{1}^{\mathrm{e}} \frac{\log \mathrm{u}}{\mathrm{u}(\mathrm{u}+1)} \mathrm{du}\)
Using change of variable
\(\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\int_{1}^{\mathrm{e}} \frac{(\log t)}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt} \tag{iii}\)
From equation (i) and (iii), we get-
\(f(e)+f\left(\frac{1}{e}\right)=\int_{1}^{e} \frac{\log t}{1+t} d t+\int_{1}^{e} \frac{\log t}{t(1+t)} \cdot d t=\int_{1}^{e} \frac{\log t}{t} d t\)
Let,
\(\log \mathrm{t} =\mathrm{v} \\ \text { Then, } \quad \frac{1}{\mathrm{t}} \mathrm{dt} =\mathrm{dv}\)
\(\mathrm{f}(\mathrm{e})+\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\int_{0}^{1} \mathrm{vdv}=\left[\frac{\mathrm{v}^{2}}{2}\right]_{0}^{1}=\frac{1}{2}\)
