Question

Let $f\left(x\right)=\frac{1}{x}\ln \left(\frac{X}{e^x}\right)$ then its primitive w.r.t. $x$ is

• A. $\frac{1}{2}{e}^x-\ln x+C$
• B. $\frac{1}{2}\ln x-e^x+C$
• C. $\frac{1}{2}{\ln }^{2}x-x+C$
• D. $\frac{e^x}{2x}+C$

Answer 1

The correct option is C $\frac{1}{2}{\ln }^{2}x-x+C$

$f\left(x\right)=\frac{1}{x}\ln \left(\frac{x}{e^x}\right)$

$=\ln \left(\frac{x^{\frac{1}{x}}}{e}\right)$

$=\frac{1}{x}\ln x-1$

So, $\int f\left(x\right)=\int (\frac{1}{x}\ln x-1)dx$

Let $\ln x=t=>\frac{1}{x}dx=dt$

$=\int tdt-x+c$

$=\frac{t^2}{2}-x+c$

$=\frac{(\ln x{)}^2}{2}-x+c$