Question

${\int }_0^1\frac{d}{dx}$ $ta{n}^{-1}\left(\frac{1}{x}\right)$

• A. $\frac{\pi }{4}$
• B. $\frac{-\pi }{4}$
• C. $\frac{-\pi }{2}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{-\pi }{4}$

Let $I={\int }_0^1\frac{d}{dx}$ ${\tan }^{-1}(\frac{1}{x})dx$
$\frac{d}{dx}$ ${\tan }^{-1}(\frac{1}{x})=\frac{1}{1+(\frac{1}{x}{)}^2}\times \frac{(-1)}{x^2}=-\frac{1}{1+x^2}$
So, $I={\int }_0^1-\frac{1}{1+x^2}dx$
$=[{\cot }^{-1}x{]}_0^1+C$
$I=-\frac{\pi }{4}$