The value of the integral ∫ _ 1/2 ^ 2 ^ - 1 x x dx is equal to

Let,

$I = {∫}_{\dfrac{1}{2}}^{2}\dfrac{\tan^{- 1}x}{x} d x . . . . . . . . \left(1\right)$

Now, let $x = \dfrac{1}{t} ⇒ dx = - \dfrac{1}{{t}^{2}}dt$

So,

$I = - {∫}_{2}^{\dfrac{1}{2}}\dfrac{\tan^{- 1}\left(\dfrac{1}{t}\right)}{\dfrac{1}{t}} \times \dfrac{1}{{t}^{2}} d t$

$⇒ I = - {∫}_{2}^{\dfrac{1}{2}}\tan^{- 1}\left(\dfrac{1}{t}\right) \times \dfrac{1}{t} d t$

$⇒ I = {∫}_{\dfrac{1}{2}}^{2}\cot^{- 1}t \times \dfrac{1}{t}dt$

$⇒ I = {∫}_{\dfrac{1}{2}}^{2}\cot^{- 1}x \times \dfrac{1}{x} d x . . . . . . . \left(2\right)$

Now adding \left(1\right) \text{ and } \left(2\right)$, we get,</p><p>$2I = {∫}_{\dfrac{1}{2}}^{2}\dfrac{1}{x}\left(\cot^{- 1}x + \tan^{- 1}x\right) d x$</p><p>$⇒ 2I = {∫}_{\dfrac{1}{2}}^{2}\dfrac{1}{x} \times \dfrac{\pi}{2} d x$</p><p>$⇒ 2I = \dfrac{\pi}{2}{\left[\ln x\right]}_{\dfrac{1}{2}}^{2}$</p><p>$⇒ 2I = \dfrac{\pi}{2}\left[\log_{e}2 - \log_{e}\dfrac{1}{2}\right]$</p><p>$⇒ 2I = \dfrac{\pi}{2}\left[2\log_{e}2\right]$</p><p>$⇒ I = \dfrac{\pi}{2}\log_{e}2$