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

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Aicharya · Accepted Answer

Suppose, $\frac{x-4}{x+2}=y \Rightarrow x-4=y(x+2)$ $ \Rightarrow x(1-y)=2 y+4 \Rightarrow x=\frac{2 y+4}{1-y} $ So, $f(y)=2\left(\frac{2 y+4}{1-y}ight)+1$ Now, $f(x)=2\left(\frac{2 x+4}{1-x}ight)+1=\frac{3 x+9}{1-x}$ $ \begin{aligned} &=\frac{3(x+3)}{1-x}=\frac{3(x-1+4)}{1-x}=-3+\frac{12}{1-x} \ &	herefore \int f(x) d x=-12 \log _e|1-x|-3 x+c \end{aligned} $

If $f\left(\frac{x-4}{x+2}\right)=2 x+1,(x \in R=\{1,-2\})$, then $\int f$ ( $x$ ) $d x$ is equal to (where…

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