Question

Let $f\left(x\right)=\frac{1}{x}In\left(\frac{x}{e^x}\right)$ then its primitive with respect to x is

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

Answer 1

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

Let I = f(x)
$\int Idx=\int \frac{1}{x}(Inx-In{e}^x)dx$
$\int \frac{Inx-x}{x}dx=[\int \frac{1}{x}Inxdx-\int \frac{1}{x}xdx]$ (Put In x =u; $\frac{1}{x}dx=du$)
= $\frac{1}{2}l{n}^{2}x-x+C$