Question

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

• A. $e^{x+2}\log \left|\frac{3x-4}{3x+4}\right|+c$
• B. $-\frac{8}{3}\log |1-x|+\frac{2}{3}x+c$
• C. $\frac{8}{3}\log |1-x|+\frac{x}{3}+c$
• D. $e^{\left[(3x-4)/(3x+4)\right]}-\frac{x^2}{2}-2x+c$

Answer 1

The correct option is B $-\frac{8}{3}\log |1-x|+\frac{2}{3}x+c$

Let $z=\left(\frac{3x-4}{3x+4}\right)$
$\therefore \frac{z+1}{z-1}=\frac{\frac{3x-4}{3x+4}}{\frac{3x-4}{3x+4}}=\frac{3x-4+3x+4}{3x-4-3x-4}=\frac{-3}{4}x$
$\therefore f\left(z\right)=\frac{4}{3}\left(\frac{z+1}{1-z}\right)+2=\frac{4z+4+6-6z}{3(1-z)}=\frac{10-2z}{3(1-z)}=\frac{8}{3(1-z)}+\frac{2}{3}$
$\therefore f\left(x\right)=\frac{8}{3(1-x)}+\frac{2}{3}$
$\therefore \int f\left(x\right)dx=\int \frac{8}{3(1-x)}dx+\int \frac{2}{3}dx=\frac{-8}{3}\log |1-x|+\frac{2}{3}x+c$