Question

The value of ${\int }_1^e(\frac{{\tan }^{-1}x}{x}+\frac{\log x}{1+x^2})dx$ is

• A. $\tan e$
• B. ${\tan }^{-1}e$
• C. ${\tan }^{-1}\left(1/e\right)$
• D. None of these

Answer 1

The correct option is B ${\tan }^{-1}e$

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

$={\int }_1^e\frac{{\tan }^{-1}x}{x}dx+{\int }_1^e\frac{\log x}{1+x^2}dx$

Using $uv$ formula for integration taking $\log x$ as first function and $\frac{1}{1+x^2}$ as second function

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

$={\tan }^{-1}e$