If f3x−43x+4=x+2, then ∫f(x)dx is

Put x=3t−43t+4=1−83t+4,dx=24(3t+4)2dtNow ∫f(x)dx=∫f3t−43t+424(3t+4)2dt=∫t+2(3t+4)2(24)dt=8∫3t+4+2(3t+4)2dt=8∫13t+4+2(3t+4)2dt=83log⁡|3t+4|−2383t+4+C1=83log⁡81−x−23(1−x)+C1=−83log⁡|1−x|+23x+C

If f3x−43x+4=x+2, then ∫f(x)dx is

Put x=3t−43t+4=1−83t+4,dx=24(3t+4)2dtNow ∫f(x)dx=∫f3t−43t+424(3t+4)2dt=∫t+2(3t+4)2(24)dt=8∫3t+4+2(3t+4)2dt=8∫13t+4+2(3t+4)2dt=83log⁡|3t+4|−2383t+4+C1=83log⁡81−x−23(1−x)+C1=−83log⁡|1−x|+23x+C

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

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

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

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

$e^{[(3 x - 4) (3 x + 4)]} - \frac{x^{2}}{2} - 2 x + c$

answer is B.

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Detailed Solution

Put $x = \frac{3 t - 4}{3 t + 4} = 1 - \frac{8}{3 t + 4}, d x = \frac{24}{(3 t + 4)^{2}} d t$

Now $\begin{array}{l} \int f (x) d x = \int f (\frac{3 t - 4}{3 t + 4}) \frac{24}{(3 t + 4)^{2}} d t \\ = \int \frac{t + 2}{(3 t + 4)^{2}} (24) d t \\ = 8 \int \frac{3 t + 4 + 2}{(3 t + 4)^{2}} d t \end{array}$

$\begin{array}{l} = 8 \int [\frac{1}{3 t + 4} + \frac{2}{(3 t + 4)^{2}}] d t \\ = \frac{8}{3} \log | 3 t + 4 | - \frac{2}{3} \frac{8}{3 t + 4} + C_{1} \\ = \frac{8}{3} \log |\frac{8}{1 - x}| - \frac{2}{3} (1 - x) + C_{1} \\ = - \frac{8}{3} \log | 1 - x | + \frac{2}{3} x + C \end{array}$