Question

If $f\left(\frac{x-4}{x+2}\right)=2x+1,(xϵR=1,-2)$ then $\int f\left(x\right)dx$ is equal to (where $C$ is a constant of integration )

Answer 1

Given $f\left(\frac{x-4}{x+2}\right)=2x+1$ ...(1)

Let $y=\frac{x-4}{x+2}$

$\Rightarrow yx+2y=x-4$

$\Rightarrow yx-x=-4-2y$

$\Rightarrow (y-1)x=-4-2y$

$\Rightarrow x=\frac{-4-2y}{y-1}$

equation (1) can be written as.

$f\left(y\right)=2(\frac{-4-2y}{y-1}+1=\frac{-8-4y+y-1}{y-1}$

$=\frac{-9-3y}{y-1}$

$\therefore f\left(y\right)=\frac{-3(3+y)}{y-1}dx.$

Now I$=\int f\left(x\right)dx=\int \frac{-3(3+x)}{x-1}dx.$

$=-3\int \frac{4-1+x}{x-1}dx$

$=-3\int \frac{x-1+4}{x-}dx$

$=-3[\int \frac{x-1}{x-1}dx+\int \frac{4dx}{x-1}]$

$=-3[\int dx+4\int \frac{dx}{x-1}]$

$=-3[x+4log(x-1\left)\right]+c$

$=-3x-12log(x-1)+c.$