If $f\left( x \right) = \int {\dfrac{{lnx}}{{1 + x}}dx} $, then $f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = $

Hint: Given the function. We have to find the integral of the function. First, find the value of $f\left( x \right)$. Then find the value of $f\left( {\dfrac{1}{x}} \right)$ by replacing the $x$ by $\dfrac{1}{x}$in the function. Then, add the functions and apply the quotient rule of logarithms to the second integral. Then, simplify the expression by canceling out the common terms. Then, replace the logarithmic function at the numerator by another variable u. Then evaluate the integral and again replace the variable by logarithmic function.

Complete step by step solution:
First, we will find the value of $f\left( x \right)$.

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

Let the function in terms of t be $f\left( t \right) = \int {\dfrac{{\ln t}}{{1 + t}}dt} $

Let $\dfrac{1}{x} = t$. Find the derivative of both sides, we get:

$\begin{gathered}
  \left( {\dfrac{1}{x}} \right) = \dfrac{{dt}}{{dx}} \\
   - \dfrac{1}{{{x^2}}}dx = dt \\
\end{gathered} $

Then, find the value of $f\left( {\dfrac{1}{x}} \right)$ by substituting the values.

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

Now, find the value of $f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$:

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

Now, apply the quotient property of logarithms to the second integration.

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

Substitute $\ln 1 = 0$ into the expression, we get:

$ \Rightarrow f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \int {\dfrac{{lnx}}{{1 + x}}dx} + \int {\dfrac{{0 - \ln x}}{{1 + \dfrac{1}{x}}}\left( { - \dfrac{1}{{{x^2}}}} \right)dx} $

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

Simplify the expression, we get:

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

Now, apply the formula $\dfrac{1}{{x\left( {1 + x} \right)}} = \dfrac{1}{x} - \dfrac{1}{{x + 1}}$ to the expression.

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

We will cancel out the common terms.

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

Substitute $u = \ln x$ into the integral and differentiate both sides, we get:
$\dfrac{{du}}{{dx}} = \dfrac{1}{x}$

Multiply $dx$ on both sides, we get:

$du = \dfrac{{dx}}{x}$

Replace $\ln x$ by $u$ and $\dfrac{{dx}}{x}$ by $du$ into the integral.

$ \Rightarrow f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \int {udu} $

Apply the power rule of integration, we get:

$ \Rightarrow f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \dfrac{{{u^2}}}{2} + C$

Replace $u$ by $\ln x$, we get:

$ \Rightarrow f\left( x \right) + f\left( {\dfrac{1}{x}} \right) = \dfrac{1}{2}{\left( {\ln x} \right)^2} + C$.

Final answer: Hence the value of $f\left( x \right) + f\left( {\dfrac{1}{x}} \right)$ is $\dfrac{1}{2}{\left( {\ln x} \right)^2} + C$.

Note: In such types of questions the quotient rule of logarithms can be applied. The quotient rule is given by $\ln \left( {\dfrac{x}{y}} \right) = \ln x - \ln y$. Also, the derivative of $\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right)$ is $\ln x$. The power rule of integration is given by $\int {{x^a}dx} = \dfrac{{{x^{a + 1}}}}{{a + 1}}$. Students may forget to add the constant of integration after solving the integral.