$\int \log (2 - 3 x) d x =$

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$\frac{1}{3} (x - 2) \log | 2 - 3 x | + 3 x + c$

$(x - \frac{2}{3}) \log | 2 - 3 x | - x + \frac{2}{3} + c$

$(x - 2) \log | 2 - 3 x | + 3 x + c$

$\frac{1}{3} (x - 2) \log | 2 - 3 x | - 3 x + c$

answer is A.

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Detailed Solution

$I = \int \log (2 - 3 x) d x$

Let $2 - 3 x = t \Rightarrow - 3 d x = d t \Rightarrow d x = \frac{- 1}{3} d t$

$= \frac{- 1}{3} \int \log t d t$

$= \frac{- 1}{3} [t \log t - t] + c (∵ \int \log x . d x = x \log x - x + c)$

$= \frac{- 1}{3} [(2 - 3 x) \log t - (2 - 3 x)] + c$

$= (x - \frac{2}{3}) \log | 2 - 3 x | - x + \frac{2}{3} + c$

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