Question

If $I={\int }_1^3({\tan }^{-1}\left(\frac{x^2-5x+6}{x^3-6x^2+12x-7}\right)+{\tan }^{-1}\left(\frac{1}{x^2-4x+4}\right))$ then the value of $\left[I\right]$ is
(where [.] represents the greatest integer function)

Answer 1

$I=\underset{1}{\overset{3}{\int }}[{\tan }^{-1}\left(\frac{{(x-2)}^2-(x-2)}{{(x-2)}^3+1}\right)+{\tan }^{-1}\left(\frac{1}{{(x-2)}^2}\right)]dx$

Let $(x-2)=t$

$=\underset{-1}{\overset{1}{\int }}({\tan }^{-1}\left(\frac{t^2-t}{1+t^3}\right)+{\tan }^{-1}\left(\frac{1}{t^2}\right))dt$

$=\underset{-1}{\overset{1}{\int }}({\tan }^{-1}\left(t^2\right)-{\tan }^{-1}t+{\tan }^{-1}\left(\frac{1}{t^2}\right))dt$

$=\underset{-1}{\overset{1}{\int }}({\tan }^{-1}\left(t^2\right)-{\tan }^{-1}\left(t\right)+{\cot }^{-1}\left(t^2\right))dt$
$=\pi$