Question

If $f\left(\frac{3x-4}{3x-4}\right) = x +2 ,x\ne - \frac{4}{3} $ and $\int f(x) dx = A \log |1 - x| + Bx + C$, then the ordered pair $(A, B)$ is equal to : (where $C$ is a constant of integration)

• A. $\left(\frac{8}{3} , \frac{2}{3}\right) $
• B. $\left( - \frac{8}{3} , \frac{2}{3}\right) $
• C. $\left( - \frac{8}{3} , - \frac{2}{3}\right) $
• D. $\left(\frac{8}{3} , - \frac{2}{3}\right) $

Answer 1

Answer: (B)

$f\left(\frac{3 x-4}{3 x-4}\right)=x+2, x \neq=\frac{4}{3}$
Let $\frac{3 x-4}{3 x+4}=t$
$3 x-4=3 t x+4 t$
$x=\frac{4 t+4}{3-3 t}+2$
$f(t)=\frac{10-2 t}{3-3 t}$
$f(t)=\frac{2 x-10}{3 x-3}$
$\int f(x) d=\int \frac{2 x-10}{3 x-3} d x$
$=\int \frac{2 x}{3 x-3} d x-10 \int \frac{d x}{3 x-3}$
$=\frac{2}{3} \int \frac{x-1}{x-1} d x+\frac{2}{3} \int \frac{d x}{x-1}-\frac{10}{3} \int \frac{d x}{x-1} $
$=\frac{2 x}{3}-\frac{8}{3} \operatorname{In}(x-1)+C$