Question

$\int e^x(logx+\frac{1}{x})$ dx is equal to

• A. $e^x+c$
• B. $e^{x}logx+c$
• C. $\frac{e^x}{x}+c$
• D. log x + c

Answer 1

Answer: (B)

$e^{x}logx+c$

$\int e^x(\log x+\frac{1}{x})dx$

As we know that,

$\int e^x(f\left(x\right)+f^{\prime }\left(x\right))dx=e^{x}f\left(x\right)+C$

$\because \frac{d}{dx}\left(\log x\right)=\frac{1}{x}$

$\therefore \int e^x(\log x+\frac{1}{x})dx=e^x\log x+C$