Question

If the function $f\left( \dfrac{x-4}{x+2} \right)=2x+1,\left( x\in \mathbb{R}\backslash \left\{ 1,-2 \right\} \right)$, then $\int{f\left( x \right)dx}$ is equal to
A. $12{{\log }_{e}}\left| 1-x \right|-3x+C$
B. $-12{{\log }_{e}}\left| 1-x \right|-3x+C$
C. $-12{{\log }_{e}}\left| 1-x \right|+3x+C$
D. $12{{\log }_{e}}\left| 1-x \right|+3x+C$

Answer 1

Hint: We first try to find the exact function for the given relation $f\left( \dfrac{x-4}{x+2} \right)=2x+1$. Assuming the value of $y=\dfrac{x-4}{x+2}$, we find out the value of x based on y. Then we put the values to find the exact function value. We change the variable and then complete the integration part. We separate the functions in two parts to integrate them. We use the partial form to find the final solution.

Complete step by step solution:
We first need to find the actual function in its simplest form.
 We have been given that $f\left( \dfrac{x-4}{x+2} \right)=2x+1,\left( x\in \mathbb{R}\backslash \left\{ 1,-2 \right\} \right)$.
We take $y=\dfrac{x-4}{x+2}$. From this relation we try to find the value of the $x$.
$\begin{align}
  & y=\dfrac{x-4}{x+2} \\
 & \Rightarrow xy+2y=x-4 \\
\end{align}$
We take all the $x$ in one side and get
\[\begin{align}
  & xy+2y=x-4 \\
 & \Rightarrow x\left( y-1 \right)=-2y-4 \\
 & \Rightarrow x=\dfrac{-2y-4}{\left( y-1 \right)}=\dfrac{2\left( y+2 \right)}{\left( 1-y \right)} \\
\end{align}\]
We put the values in the function $f\left( \dfrac{x-4}{x+2} \right)=2x+1$ to get
$\begin{align}
  & f\left( \dfrac{x-4}{x+2} \right)=2x+1 \\
 & \Rightarrow f\left( y \right)=2\left( \dfrac{2\left( y+2 \right)}{\left( 1-y \right)} \right)+1 \\
 & \Rightarrow f\left( y \right)=\dfrac{4\left( y+2 \right)}{\left( 1-y \right)}+1=\dfrac{3y+9}{\left( 1-y \right)} \\
\end{align}$
Simplifying and changing the variable we get $f\left( x \right)=\dfrac{3x+9}{1-x}$.
Now we have to integrate the function to get
\[\int{f\left( x \right)dx}=\int{\dfrac{3\left( x+3 \right)}{1-x}dx}\].
We change the numerator as \[3\left( x+3 \right)=-3\left( 1-x \right)+12=12-3\left( 1-x \right)\].
So, \[\int{f\left( x \right)dx}=\int{\dfrac{12-3\left( 1-x \right)}{1-x}dx}\].
Now we separate the functions and get
\[\begin{align}
  & \int{\dfrac{12-3\left( 1-x \right)}{1-x}dx} \\
 & =\int{\dfrac{12}{1-x}dx}+\int{\dfrac{-3\left( 1-x \right)}{1-x}dx} \\
 & =-12\int{\dfrac{\left( -dx \right)}{1-x}}-3\int{dx} \\
 & =-12\int{\dfrac{d\left( 1-x \right)}{1-x}}-3\int{dx} \\
 & =-12\log \left| 1-x \right|-3x+c \\
\end{align}\]
Therefore, the correct option is B.

Note: A differential 1-form is integrated along an oriented curve as a line integral and it is integrated just like a surface integral. because the square whose first side is $dx$ and second side is $dy$, to be regarded as having the opposite orientation as the square whose first side is $dx$ and whose second side is $dy$.