Question

The value of the integral ${\int }_1^2{e}^x(lo{g}_{e}X+\frac{x+1}{x})dx$ 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

The correct option is C $e^2(1+lo{g}_{e}2)-e$

Given : ${\int }_1^2{e}^x({\log }_{e}x+(x+\frac{1}{x}))dx$

${\int }_1^2{e}^x({\log }_{e}x+(x+\frac{1}{x}))dx{\int }_1^2{e}^x({\log }_{e}x+(x+\frac{1}{x}))dx={\int }_1^2{e}^x(\ln x+\frac{1}{x})dx+{\int }_1^2{e}^{x}dx={\left|e^x\ln x\right|}_1^2+{\left|e^x\right|}_1^2=\int e^x\left(f\right(\left(x\right))+f^{\prime }(x)dx=e^{x}f\left(x\right)+C=e^2\ln 2-0+e^2-e^1=e^2(\ln 2+1)-e+C$

Hence the correct answer is $e^2(\ln 2+1)-e$