If fleft(frac{x-4}{x-2}right)=2x+1,x in R-{1, | vedclass

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(A) Given $f\left(\frac{x-4}{x-2}\right)=2x+1$. Let $y = \frac{x-4}{x-2}$. Then $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$. So,$\frac{2}{x-2} = 1-y$,

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$5x-4 \log |x-1|+c$,where $c$ is constant of integration.

Vedclass · Answer

(A) Given $f\left(\frac{x-4}{x-2}\right)=2x+1$. Let $y = \frac{x-4}{x-2}$. Then $y = \frac{x-2-2}{x-2} = 1 - \frac{2}{x-2}$. So,$\frac{2}{x-2} = 1-y$,which implies $x-2 = \frac{2}{1-y} = \frac{-2}{y-1}$. Thus,$x = 2 - \frac{2}{y-1} = \frac{2y-2-2}{y-1} = \frac{2y-4}{y-1}$. Now,$f(y) = 2x+1 = 2\left(\frac{2y-4}{y-1}\right) + 1 = \frac{4y-8+y-1}{y-1} = \frac{5y-9}{y-1} = \frac{5(y-1)-4}{y-1} = 5 - \frac{4}{y-1}$. Therefore,$f(x) = 5 - \frac{4}{x-1}$. Now,$\int f(x) dx = \int \left(5 - \frac{4}{x-1}\right) dx = 5x - 4 \log |x-1| + c$.

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

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