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Q.
The value of definite integral $\int\limits_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{x^2-x+1} d x$ is equal to
Integrals
Solution:
$I=\int\limits_{\frac{1}{2}}^2 \frac{\tan ^{-1} x}{x^2-x+1} d x$......(1)
put $x=\frac{1}{t}$
$\therefore I =\int\limits_{\frac{1}{2}}^2 \frac{\tan ^{-1} \frac{1}{ t }}{ t ^2- t +1} dt$....(2)
Now, (1) + (2) gives
$2 I =\frac{\pi}{2} \int\limits_{\frac{1}{2}}^2 \frac{ dx }{\left( x -\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2} $
$\left.\therefore I =\frac{\pi}{4}\left(\frac{2}{\sqrt{3}}\right) \tan ^{-1}\left(\frac{2 x -1}{\sqrt{3}}\right)\right]_{\frac{1}{2}}^2 $
$=\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}-0\right)=\frac{\pi^2}{6 \sqrt{3}} \Rightarrow A$