Question

If $f\left[\frac{(3x-4)}{(3x+4)}\right]=x+2,$ then $\int f\left(x\right)dx=?$

Answer 1

Finding the value of $\mathbf{\int }\mathit{f}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathit{d}\mathit{x}\mathbf{}$ :

Given that $f\left[\frac{(3x-4)}{(3x+4)}\right]=x+2,$

Let $\frac{(3x-4)}{(3x+4)}=\alpha$

Apply Componendo-dividendo rule

We get $\frac{(3x-4)+(3x+4)}{(3x-4)-(3x+4)}=\frac{\alpha +1}{\alpha -1}$

$$\begin{gathered} \Rightarrow \frac{6x}{-8}=\frac{\alpha +1}{\alpha -1} \\ \Rightarrow x=-\frac{4}{3}\left(\frac{\alpha +1}{\alpha -1}\right)....\left(i\right) \end{gathered}$$

Now $x+2=-\frac{4}{3}\left(\frac{\alpha +1}{\alpha -1}\right)+2$ ,Since $x+2=f\left(\alpha \right)$

$\Rightarrow$ $f\left(\alpha \right)=\frac{-4\alpha -4+2\times 3(\alpha -1)}{3\alpha -3}$

$\Rightarrow$ $f\left(\alpha \right)=\frac{2\alpha -10}{3\alpha -3}$

$\Rightarrow$ $f\left(\alpha \right)=\frac{10-2\alpha }{3-3\alpha }$

Similarly $f\left(x\right)=\frac{10-2x}{3-3x}$

Now ,

$$\begin{gathered} \int f\left(x\right)=\int \frac{10-2x}{3-3x}dx \\ =\frac{2}{3}\int \frac{5-x}{1-x}dx \\ =\frac{2}{3}\int (\frac{4}{1-x}+1)dx \\ =\frac{2}{3}\int \frac{4}{1-x}dx+\frac{2}{3}dx \\ =-\frac{8}{3}\log (1-x)+\frac{2}{3}x+C \\ =\frac{-8}{3}\log \left|1-x\right|+\frac{2}{3}x+C \end{gathered}$$

Hence, the value of $\mathbf{\int }\mathit{f}\mathbf{}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathit{d}\mathit{x}\mathbf{}\mathbf{=}$ $\frac{\mathbf{-}\mathbf{8}}{\mathbf{3}}\mathit{l}\mathit{o}\mathit{g}\left|1-x\right|\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\mathit{x}\mathbf{}\mathbf{+}\mathit{C}$.