Question

${\int }_1^e(\frac{{\tan }^{-1}x}{x}+\frac{\ln x}{1+x^2})dx$ is equal to

• A. $\frac{1}{e}{\tan }^{-1}e$
• B. ${\tan }^{-1}e$
• C. $e{\tan }^{-1}\left(\frac{1}{e}\right)$
• D. ${\tan }^{-1}\left(\ln e\right)$

Answer 1

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

${\int }_1^e{\tan }^{-1}x.\frac{1}{x}dx+{\int }_1^e\frac{\ln x}{1+x^2}dx$
$I$ $II$
$=({\tan }^{-1}x.\ln x{)}_1^e-{\int }_1^e\frac{1}{(1+x^2)}\ln xdx+{\int }_1^e\frac{\ln xdx}{(1+x^2)}$
$=({\tan }^{-1}e.\ln e)-{\tan }^{-1}1.\ln 1$
$={\tan }^{-1}e$