Question

If $f\left(\frac{2x+3}{3x+5}\right)=x+4x\ne -\frac{5}{3},\frac{2}{3}$ and $\int f(x)dx=Ax+B\ln |3x-2|+C$, then $3B-A=$

• A. $\frac{64}{9}$
• B. $\frac{-52}{21}$
• C. $\frac{-10}{3}$
• D. $-\frac{8}{3}$

Answer 1

Answer: (D)

We have, $f\left(\frac{2x+3}{3x+5}\right)=x+4$
Put, $\frac{2x+3}{3x+5}=y$
$\Rightarrow 2x+3=3xy+5y$
$\Rightarrow x=\frac{5y-3}{2-3y}$
$\therefore f(y)=\frac{5y-3}{2-3y}+4=\frac{5y-3+8-12y}{2-3y}$
$\Rightarrow f(y)=\frac{7y-5}{3x-2}$
$\therefore f(x)=\frac{7x-5}{3x-2}$
$\Rightarrow \int f(x)dx=\int \frac{7x-5}{3x-2}dx$
Put, $3x-2=t\Rightarrow x=\frac{t+2}{3}$
$dx=\frac{dt}{3}=\frac{1}{3}\int \frac{7\left(\frac{t+2}{3}\right)-5}{t}dt$
$=\frac{1}{9}\int \frac{7t-1}{t}dt=\frac{7}{9}\int dt-\frac{1}{3}\int \frac{dt}{t}$
$=\frac{7}{9}t-\frac{1}{9}\log t+C$
$=\frac{7}{9}(3x-2)-\frac{1}{9}\log |3x-2|+C$
$=\frac{7}{3}x-\frac{1}{9}\log |3x-2|+C$
Here, $A=\frac{7}{3},B=-\frac{1}{9}$
$\therefore 3B-A=3\left(-\frac{1}{9}\right)-\frac{7}{3}=-\frac{8}{3}$