For x > 0,
Question:

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

  1. 1

  2. -1

  3. $\frac{1}{2}$

  4. 0


Correct Option: , 3

Solution:

$f(x)=\int_{i}^{x} \frac{\log _{e} t}{(1+t)} d t$

$f\left(\frac{1}{x}\right)=\int_{1}^{1 / x} \frac{\ell \mathrm{nt}}{1+t} \mathrm{dt}$, let $\mathrm{t}=\frac{1}{\mathrm{y}}$

$=+\int_{1}^{x} \frac{\ell \text { ny }}{1+y} \cdot \frac{y}{y^{2}} d y$

$=\int_{1}^{x} \frac{\ell n y}{y(1+y)} d y$

hence

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

$=\frac{1}{2} \ell \mathrm{n}^{2}(\mathrm{x})$

so $f(\mathrm{e})+f\left(\frac{1}{\mathrm{e}}\right)=\frac{1}{2}$

..(3)

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