If f(x)=∫1x log⁡t1+tdt, then f(x)+f1x is equal to

A
${(\log_{c} x)}^{2}$
B
$\frac{2}{3} \log_{e} x$
C
$\frac{1}{2} \log_{e} x$
D
$\frac{1}{2} {(\log_{e} x)}^{2}$

Solution:

We have,

$\begin{array}{l} f (x) = \int_{1}^{x} \frac{\log t}{1 + t} d t \\ f (\frac{1}{x}) = \int_{1}^{1 / x} \frac{\log t}{1 + t} d t \end{array}$

$\Rightarrow f (\frac{1}{x}) = \int_{1}^{x} \frac{\log (\frac{1}{u})}{1 + \frac{1}{u}} \times (- \frac{1}{u^{2}}),$ where $t = \frac{1}{u}$

$\begin{array}{l} \Rightarrow f (\frac{1}{x}) = \int_{1}^{x} \frac{\log u}{u (1 + u)} d u \\ \Rightarrow f (\frac{1}{x}) = \int_{1}^{x} \frac{\log t}{t (1 + t)} d t \\ f (x) + f (\frac{1}{x}) \\ = \int_{1}^{x} \frac{\log t}{1 + t} d t + \int_{1}^{x} \frac{\log t}{t (1 + t)} d t \end{array}$

$= \int_{1}^{x} \frac{\log t}{t} d t = \int_{1}^{x} \log t d (\log t) = {[\frac{(\log t)^{2}}{2}]}_{1}^{x} = \frac{{(\log_{c} x)}^{2}}{2}$

If $f (x) = \int_{1}^{x} \frac{\log t}{1 + t} d t,$ then $f (x) + f (\frac{1}{x})$ is equal to

Solution:

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