If f((x - 4)/(x + 2)) = 2x + 1, (x ∈ R - \ 1, - 2\ ) , then ∫...

We have, $f\left(\dfrac{x - 4}{x + 2}\right) = 2x + 1$

Let $\dfrac{x - 4}{x + 2} = X ⇒ xX + 2X = x - 4 ⇒ x$\left(X - 1\right) = - 4 - 2X ⇒ x = \dfrac{2\left(2 + X\right)}{\left(1 - X\right)}

Putting $x = \dfrac{2\left(2 + X\right)}{\left(1 - X\right)}$ in the given functional equation, we get

$⇒ f\left(X\right) = 2 \times 2\left\{\dfrac{2 + X}{1 - X}\right\} + 1 = \dfrac{8 + 4X + 1 - X}{1 - X} = \dfrac{3X + 9}{1 - X} = \dfrac{3\left(X + 3\right)}{\left(1 - X\right)}$

Replacing $X \to x$, we get

$f\left(x\right) = \dfrac{3\left(x + 3\right)}{\left(1 - x\right)}$

$⇒ ∫ f\left(x\right)dx = 3 ∫ \left(\dfrac{x + 3}{1 - x}\right)dx = 3 ∫ \dfrac{4 - \left(1 - x\right)}{1 - x}dx = 3\left\{∫ \dfrac{4}{1 - x}dx - ∫ dx\right\}$

$= - 12\ln \left|1 - x\right| - 3x + C$, where $C$ is the constant of integration.