Question

$\int \frac{\text{ln}\left(\text{e}{\text{x}}^{\text{x}}\right)}{\text{x}{\text{e}}^{\text{x}}{\left(\text{lnx}\right)}^{\text{2}}}\text{dx}\text{is}$

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

Answer 1

The correct option is C $-\frac{1}{{\text{e}}^{\text{x}}\text{lnx}}\text{+ c}$

Now,

$\int \frac{\text{ln}\left(\text{e}{\text{x}}^{\text{x}}\right)}{\text{x}{\text{e}}^{\text{x}}{\left(\text{lnx}\right)}^{\text{2}}}\text{dx}$

$=\int \frac{(1+x\ln x)}{x{e}^x(\ln x{)}^2}dx$

$=\int \frac{(e^x+x{e}^x\ln x)}{x{e}^{2x}(\ln x{)}^2}dx$

$=\int \frac{(\frac{e^x}{x}+e^x\ln x)}{(e^x\ln x{)}^2}dx$

$=\int \frac{d\left(e^x\ln x\right)}{(e^x\ln x{)}^2}$

$=-\frac{1}{e^x\ln x}+c$ [Where $c$ is integrating constant]