Question

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

• A. e^x+2 ln $\left| \frac{3x - 4}{3x + 4} \right|$ + c
• B. $- \frac{8}{3}$ln |(1 – x)| + $\frac{2}{3}$x + c
• C. $\frac{8}{3}$ln |x – 1| + $\frac{x}{3}$ + c
• D. None of these

Answer 1

Answer: (B)

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

Answer 2

$\frac { 3 x - 4 } { 3 x + 4 }$ = t

3x – 4 = 3xt + 4t

x = $\frac{4t + 4}{3(1 - t)}$

f(t) = $\frac{4t + 4}{3(1 - t)}$ + 2

f(x) = $\frac{4x + 4}{3(1 - x)}$ + 2 = $\frac{4(x - 1) + 8}{3(1 - x)}$ + 2

f(x) = 2 – $\frac{4}{3}$ – $\frac{8}{3(x - 1)}$ = $\frac{2}{3} - \frac{8}{3(x - 1)}$

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