Question

Solve:-
$\underset{-1}{\overset{1}{\int }}\frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)dx$

Answer 1

$Since,\frac{d}{dx}{\tan }^{-1}\frac{1}{x}=\frac{1}{1+{\left(\frac{1}{x}\right)}^2}=\frac{x^2}{x^2+1}$

${\int }_{-1}^1\frac{d}{dx}{\tan }^{-1}\frac{1}{x}dx={\int }_{-1}^1\frac{x^2}{x^2+1}dx$

$={\int }_{-1}^1\frac{x^2+1-1}{x^2+1}dx={\int }_{-1}^1(\frac{x^2+1}{x^2+1}-\frac{1}{x^2+1})dx$

$=2{\int }_0^1(1-\frac{1}{x^2+1})dx$

( using definite integral property when f(-x)=f(x) then ${\int }_{-a}^{a}dx=2{\int }_0^{a}dx$)

$=2{(x-{\tan }^{-1}x)}_0^1=2(1-\frac{\Pi }{4})(Ans.)$