If $\mathrm{f}\left(\frac{x-4}{x-2}\right)=2 x+1, x \in \mathbb{R}-\{1,-2\}$, then $\int \mathrm{f}(x)…
If $\mathrm{f}\left(\frac{x-4}{x-2}\right)=2 x+1, x \in \mathbb{R}-\{1,-2\}$, then $\int \mathrm{f}(x) \mathrm{d} x$ is equal to
- $5 x-4 \log (x-1)+\mathrm{c}$, where c is constant of integration.
- $x-4 \log (x-1)+c$, where c is constant of integration.
- $5 x+4 \log (x-1)+\mathrm{c}$, where c is constant of integration.
- $5 x+\log (x-1)+\mathrm{c}$, where c is constant of integration.
Solution
$\begin{aligned} & \mathrm{f}\left(\frac{x-4}{x-2}\right)=2 x+1 \\ & \text { Let } \frac{x-4}{x-2}=y \\ & \frac{x-2-2}{x-2}=y \\ & 1-\frac{2}{x-2}=y \\ & \frac{-2}{x-2}=y-1 \\ & \frac{-2}{y-1}=x-2 \\ & \Rightarrow x=\frac{-2}{y-1}+2 \\ & \Rightarrow 2 x=\frac{-4}{y-1}+4 \\ & \Rightarrow 2 x+1=\frac{-4}{y-1}+5\end{aligned}$
$\begin{aligned}
& \therefore \quad \mathrm{f}(y)=\frac{-4}{y-1}+5 \\
& \Rightarrow \mathrm{f}(x)=\frac{-4}{x-1}+5 \\
& \therefore \quad \int \mathrm{f}(x) \mathrm{d} x=\int \frac{-4}{x-1}+5 \\
& =-4 \log (x-1)+5 x+c \\
& =5 x-4 \log (x-1)+c
\end{aligned}$
Asked in: MHT CET 2024 (03 May Shift 2)
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