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**Hint:**In this question, we will first find $f\left( \dfrac{1}{x} \right)$ using substitution in $f\left( x \right)$. Then we will add both $f\left( x \right)$ and $f\left( \dfrac{1}{x} \right)$ to get simple expression. Then we will evaluate the integral using substitution again and hence find our answer. Only basic integration formula $\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c$ will be used in this question.

**Complete step-by-step solution**We are given $f\left( x \right)=\int{\dfrac{\ln x}{1+x}dx}....................(1)$

We need to find $f\left( \dfrac{1}{x} \right)$. So we will first consider integral in terms of variable $t$ to avoid confusion.

$f\left( t \right)=\int{\dfrac{\ln t}{1+t}dt}.....................(2)$

To find the value of $f\left( \dfrac{1}{x} \right)$, let us substitute the value of $\dfrac{1}{x}$ by $t$.

$\therefore \dfrac{1}{x}=t....................(3)$

Taking derivative both sides, we get –

$-\dfrac{1}{{{x}^{2}}}dx=dt....................(4)$

Putting (3) and (4) in equation (2), we get –

$f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln \dfrac{1}{x}}{1+\dfrac{1}{x}}}\left( -\dfrac{1}{{{x}^{2}}} \right)dx$

Taking LCM in denominator and simplifying, we get –

$f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln \dfrac{1}{x}.x}{1+x}}\left( -\dfrac{1}{{{x}^{2}}} \right)dx$

As we know, $\ln \dfrac{1}{x}=-\ln x$. Hence, we get –

$\begin{align}

& f\left( \dfrac{1}{x} \right)=-\int{\dfrac{-\ln x}{1+x}}\left( \dfrac{1}{x} \right)dx \\

& f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln x}{x\left( 1+x \right)}}dx.......................(5) \\

\end{align}$

Now adding (1) and (5), we get –

$f\left( x \right)+f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln x}{\left( 1+x \right)}}dx+\int{\dfrac{\ln x}{x\left( 1+x \right)}}dx$

Combining the integral, we have –

$f\left( x \right)+f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln x}{\left( 1+x \right)}}dx+\dfrac{\ln x}{x\left( 1+x \right)}dx$

Taking $\dfrac{\ln x}{\left( 1+x \right)}$ common, we get –

$\begin{align}

& f\left( x \right)+f\left( \dfrac{1}{x} \right)=\int{\dfrac{\ln x}{\left( 1+x \right)}}\left( 1+\dfrac{1}{x} \right)dx \\

& =\int{\dfrac{\ln x}{\left( 1+x \right)}}\left( \dfrac{1+x}{x} \right)dx \\

& =\int{\dfrac{\ln x}{\left( 1+x \right)}}\left( \dfrac{1+x}{x} \right)dx \\

& =\int{\dfrac{\ln x}{x}}dx......................(6) \\

\end{align}$

Now we just have to evaluate this integral to find the required result. For this, we will use substitution.

Let $\ln x=a……………………….......(7)$

Differentiating both sides, we get –

$\dfrac{1}{x}dx=da$

So, equation (6) becomes $\int{a~da}$

Solving integration, we get –

$\dfrac{{{a}^{2}}}{2}+c$

Putting the value of $a$ from equation (7), we get –

$\dfrac{{{\left( \ln x \right)}^{2}}}{2}+c$

**Hence, $f\left( x \right)+f\left( \dfrac{1}{x} \right)=\dfrac{{{\left( \ln x \right)}^{2}}}{2}+c$. Therefore, option (b) is the correct answer.**

**Note:**Students should be careful while substituting values to solve integral. Do not forget to change $dx$ also. Also, try to combine the full equation into a single integral for solving sum easily. Do not try to solve the integral of $f\left( x \right)$ and $f\left( \dfrac{1}{x} \right)$ separately and then adding. Also, don’t forget to add constant after solving integration.

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