Question

Let $y=y\left(x\right),y\left(1\right)=1andy\left(e\right)=e^2$ . Consider
$J=\int \frac{x+y}{xy}dy$, $I=\int \frac{x+y}{x^2}dx$, $J-I=g\left(x\right)$ and g(1) = 1, then the value of g(e) is

• A. $2e+1$
• B. $e+1$
• C. $e^2$-e+1
• D. $e^2$+e-1

Answer 1

The correct option is A $2e+1$

$I=\int \frac{x+y}{xy}dy;I=\int \frac{x+y}{x^2}dx$

$I=\int (\frac{1}{y}+\frac{1}{x})dy;I=\int (\frac{1}{x}+\frac{y}{x^2})dx$

$I\Rightarrow Iny+\frac{y}{x};I=Inx-\frac{y}{x}$

$y\left(x\right)=J-I$

$\Rightarrow Iny-Inx+\frac{2y}{x}$

$g\left(x\right)\Rightarrow In\left(y/x\right)+\frac{2y}{x}=In\left(\frac{y\left(x\right)}{2}\right)+\frac{2y\left(x\right)}{x}$

$g\left(1\right)=Iny+2y+1$

$g\left(2\right)=In\left|\frac{e^2}{e}\right|+\frac{2e^2}{e}$

$\Rightarrow 2e+1$