If f(3x−43x+4)=x+2, x≠−43, then ∫f(x)dx=

Put 3x−43x+4=t sothat x=4t+43−3tthen f(x)=2x−103x−3∫f(x)dx=23∫(1−4x−1)dx

If f(3x−43x+4)=x+2, x≠−43, then ∫f(x)dx=

Put 3x−43x+4=t sothat x=4t+43−3tthen f(x)=2x−103x−3∫f(x)dx=23∫(1−4x−1)dx

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

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$e^{x + 2} \ln | \frac{3 x - 4}{3 x + 4} | + c$

$\frac{- 8}{3} \ln | 1 - x | + \frac{2}{3} x + c$

$\frac{8}{3} \ln | x - 1 | + \frac{x}{3} + c$

$\frac{- 8}{3} \ln | 1 - x | - \frac{2}{3} x + c$

answer is B.

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Detailed Solution

Put $\frac{3 x - 4}{3 x + 4} = t s o$
that $x = \frac{4 t + 4}{3 - 3 t}$
then $f (x) = \frac{2 x - 10}{3 x - 3}$
$\int f (x) d x = \frac{2}{3} \int (1 - \frac{4}{x - 1}) d x$

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