# For x>0

Question:

For $x>0$, if $f(x)=\int_{1}^{x} \frac{\log _{e} t}{(1+t)} d t$, then $f(e)+f\left(\frac{1}{\epsilon}\right)$ is equal to :

1. (1) $\frac{1}{2}$

2. (2) $-1$

3. (3) 1

4. (4) 0

Correct Option: 1

Solution:

$f(e)+f\left(\frac{1}{e}\right)=\int_{1}^{e} \frac{\ell n t}{1+t} d t+\int_{1}^{1 / e} \frac{\ell n t}{1+t} d t=I_{1}+I_{2}$

$I_{2}=\int_{1}^{1 / e} \frac{\text { ent }}{1+t} d t \quad$ put $t=\frac{1}{z}, d t=-\frac{d z}{z^{2}}$

$=\int_{1}^{e}-\frac{\ell n z}{1+\frac{1}{3}} \times\left(-\frac{d z}{z^{2}}\right)=\int_{1}^{e} \frac{\ell n z}{z(z+1)} d z$

$\mathrm{f}(\mathrm{e})+\mathrm{f}\left(\frac{1}{\mathrm{e}}\right)=\int_{1}^{e} \frac{\ell \mathrm{th}}{1+t} \mathrm{dt}+\int_{1}^{e} \frac{\ell \mathrm{ut}}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}=\int_{1}^{e} \frac{\ell \mathrm{nt}}{1+t}+\frac{t \mathrm{ut}}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}$

$\left.=\int_{1}^{e} \frac{f \mathrm{t} t}{\mathrm{t}} \mathrm{dt}\left\{\ln \mathrm{t}=\mathrm{u}, \frac{1}{\mathrm{t}} \mathrm{dt}\right\}_{\left.\mathrm{u}^{2}\right|^{1}}\right\}$\

$=\mathrm{du}=\int_{0}^{1} \mathrm{udu}=\left.\frac{\mathrm{u}^{2}}{2}\right|_{0} ^{1}=\frac{1}{2}$