Question

The integral ${\int }_1^2{e}^x·x^x(2+{\log }_e\left(x\right))dx$ equals

• A. $e(4e-1)$
• B. $e(4e+1)$
• C. $4e^2-1$
• D. $e(2e-1)$

Answer 1

The correct option is A

$e(4e-1)$

Explanation for correct option

Given integral${\int }_1^2{e}^x·x^x(2+{\log }_e\left(x\right))dx$

Put $e^x·x^x=t$

Since upper limit $=e^2·2^2,$ Lower limit $=e$

$$\begin{gathered} (e^x·x^x+e^x{x}^x(1+\ln \left(x\right)\left)\right)dx=dt \\ {e}^x·x^x(2+\ln \left(x\right))dx=dt \\ {\int }_e^{4e^2}dt={{\left[t\right]}_e}^{4e^2} \\ =4e^2-e \\ =e(4e-1) \end{gathered}$$

Hence, option A is correct .