Question

The value of integral ${\int }_{-1}^3({\tan }^{-1}\left(\frac{x}{1+x^2}\right)+{\tan }^{-1}\left(\frac{x^2+1}{x}\right))dx$

Answer 1

Using the formula , $ta{n}^{-1}x+ta{n}^{-1}y$ = $ta{n}^{-1}\frac{x+y}{1-xy}$

${\int }_{-1}^3$ ($ta{n}^{-1}\frac{x}{1+x^2}$ + $ta{n}^{-1}\frac{x^2+1}{x}$)

${\int }_{-1}^3$ $ta{n}^{-1}\frac{\frac{x^2+(x^2+1{)}^2}{x(1+x^2)}}{1-1}$ = ${\int }_{-1}^3$ $\frac{\pi }{2}dx$ = $2\pi$