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

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

1. A

${e}^{x+2}{\mathrm{log}}_{e}\left|\frac{3x-4}{3x+4}\right|$

2. B

$-\frac{8}{3}{\mathrm{log}}_{e}|1-x|+\frac{2}{3}x+C$

3. C

$\frac{8}{3}{\mathrm{log}}_{t}|x-1|+\frac{x}{3}+C$

4. D

none of the

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### Solution:

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

Let $\frac{3x-4}{3x+4}=\alpha$

$\begin{array}{l}⇒f\left(\alpha \right)=\frac{2}{3}\left(\frac{\alpha -5}{\alpha -1}\right)\\ ⇒f\left(\alpha \right)=\frac{2}{3}\left(\frac{\alpha -1-4}{\alpha -1}\right)=\frac{2}{3}\left(1-\frac{4}{\alpha -1}\right)=\frac{2}{3}-\frac{8}{3\left(\alpha -1\right)}\\ ⇒f\left(x\right)=\frac{2}{3}-\frac{8}{3\left(x-1\right)}\\ \therefore \int f\left(x\right)dx=\int \left\{\frac{2}{3}-\frac{8}{3\left(x-1\right)}\right\}dx\\ ⇒\int f\left(x\right)dx=\frac{2}{3}x-\frac{8}{3}{\mathrm{log}}_{e}|x-1|+C\end{array}$

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