Question

${\int }_1^2\left(\frac{1+x\log x}{x}\right)e^{x}d{x}_{=}$

• A. $e^2$ $log2$
• B. $elog2$
• C. $\frac{1}{2}$ log2
• D. $\frac{e^2}{2}$ log2

Answer 1

The correct option is A $e^2$ $log2$

${\int }_1^2\left(\frac{1+x\log x}{x}\right)e^{x}dx$

${\int }_1^2(\frac{1}{x}+\log x)e^{x}dx$ $\int e^x\left[f\right(x)+f^{\prime }(x\left)\right]dx=e^{x}f\left(x\right)$
$={\int }_1^2(e^x\log x+c)$
$=e^2\log 2-e\log 1$
$=e^2\log 2-0$
${\int }_1^2\left(\frac{1+x\log x}{x}\right)e^{x}dx=e^2\log 2$