Question

If we have a function as $f\left( x \right)=\int{\dfrac{\ln x}{1+x}dx}$ then $f\left( x \right)+f\left( \dfrac{1}{x} \right)=$
 $\begin{align}
  & \left( a \right){{\left( \ln x \right)}^{2}}+c \\
 & \left( b \right)\dfrac{1}{2}{{\left( \ln x \right)}^{2}}+c \\
 & \left( c \right)\dfrac{1}{2}{{\left( x\ln x \right)}^{2}}+c \\
 & \left( d \right)None~of~these \\
\end{align}$

Answer 1

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.