Question

The value of ${\int }_1^e\frac{1+x^2\ln x}{x+x^2\ln x}dx$ is

• A. $e$
• B. ln$\left(1+e\right)$
• C. $e+$In$\left(1+e\right)$
• D. $e-$ In $\left(1+e\right)$

Answer 1

Answer: (D)

$I={\int }_1^e\left(\frac{1}{x}+1\right)dx-{\int }_1^e\frac{1+\ln x}{1+x\ln x}dx$
$=[\ln x+x{]}_1^e-[\ln (1+x\ln x){]}_1^e$
$=e-\ln (1+e)$