If f3x−43x+4=x+2, then ∫f(x)dx is equal to

A
$e^{x + 2} \log_{e} |\frac{3 x - 4}{3 x + 4}|$
B
$- \frac{8}{3} \log_{e} | 1 - x | + \frac{2}{3} x + C$
C
$\frac{8}{3} \log_{t} | x - 1 | + \frac{x}{3} + C$
D
none of the

Solution:

We have, $f (\frac{3 x - 4}{3 x + 4}) = x + 2$

Let $\frac{3 x - 4}{3 x + 4} = α$

$\begin{array}{l} \Rightarrow \frac{(3 x - 4) + (3 x + 4)}{(3 x - 4) - (3 x + 4)} = \frac{α + 1}{α - 1} \\ \Rightarrow \frac{6 x}{- 8} = \frac{α + 1}{α - 1} \\ \Rightarrow x = - \frac{4}{3} (\frac{α + 1}{α - 1}) \\ \Rightarrow x + 2 = - \frac{4 α + 4}{3 α - 3} + 2 = - \frac{- 4 α - 4 + 6 α - 6}{3 α - 3} = \frac{2 α - 10}{3 α - 3} \\ ∴ f (\frac{3 x - 4}{3 x + 4}) = x + 2 \\ \Rightarrow f (α) = \frac{2 α - 10}{3 α - 3} \end{array}$

$\begin{array}{l} \Rightarrow f (α) = \frac{2}{3} (\frac{α - 5}{α - 1}) \\ \Rightarrow f (α) = \frac{2}{3} (\frac{α - 1 - 4}{α - 1}) = \frac{2}{3} (1 - \frac{4}{α - 1}) = \frac{2}{3} - \frac{8}{3 (α - 1)} \\ \Rightarrow f (x) = \frac{2}{3} - \frac{8}{3 (x - 1)} \\ ∴ \int f (x) d x = \int \{\frac{2}{3} - \frac{8}{3 (x - 1)}\} d x \\ \Rightarrow \int f (x) d x = \frac{2}{3} x - \frac{8}{3} \log_{e} | x - 1 | + C \end{array}$

If $f (\frac{3 x - 4}{3 x + 4}) = x + 2$ , then $\int f (x) d x$ is equal to

Solution:

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