Question

The value of the integral
$\underset{1}{\overset{2}{\int }}{e}^x\left(lo{g}_{e}x+\frac{x+1}{x}\right)dx$ is

• A. $e^2(1+lo{g}_{e}2)$
• B. $e^2-e$
• C. $e^2(1+lo{g}_{e}2)-e$
• D. $e^2-e(1+lo{g}_{e}2)$

Answer 1

Answer: (C)

Let $/={\int }_1^2{e}^x\left({\log }_{\theta }x+\frac{x+1}{x}\right)dx$
$\Rightarrow I={\int }_1^2\left(e^x\cdot {\log }_{e}x+e^x+\frac{e^x}{x}\right)dx$
$\Rightarrow I={\int }_1^2{e}^x{\log }_{\theta }xdx+{\int }_1^2{e}^{x}dx+{\int }_1^2\frac{e^x}{x}dx$
$\Rightarrow I={\int }_1^2{e}^x{\log }_{e}xdx+{\left[e^x\right]}_1^2+[e^x{\log }_{\theta }{x}_1^2$
$-{\int }_1^2{e}^x{\log }_{\theta }xdx$
$\Rightarrow I=\left(e^2-e^1\right)+\left(e^2{\log }_{\theta }2-0\right)$
$\Rightarrow e^2\left(1+{\log }_{\theta }2\right)-e$