Area bounded by the curve $xy-3x-2y-10=0,$ x-axis and the lines $x=3,x=4$ is

Given curve is $y(x-2)=3x+10\Rightarrow y=\frac{3x+10}{x-2}$

x-axis and the lines $x=3,x=4$

$\therefore$ required area $={\int }_3^4 ydx={\int }_3^4 \frac{3x+10}{x-2}dx$

$={\int }_3^4 \left(3+\frac{16}{x-2}\right)dx$

$=[3x+16\log (x-2){]}_3^4=3+16\log 2$ sq. units