Question

The value of definite integral $\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{{\tan }^{-1}x}{x^2-x+1}dx$ is equal to

• A. $\frac{{\pi }^2}{6\sqrt{3}}$
• B. $\frac{{\pi }^2}{3\sqrt{3}}$
• C. $\frac{{\pi }^2}{12\sqrt{3}}$
• D. $\frac{{\pi }^2}{4\sqrt{3}}$

Answer 1

Answer: (A)

$I=\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{{\tan }^{-1}x}{x^2-x+1}dx$......(1)
put $x=\frac{1}{t}$
$\therefore I=\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{{\tan }^{-1}\frac{1}{t}}{t^2-t+1}dt$....(2)
Now, (1) + (2) gives
$2I=\frac{\pi }{2}\underset{\frac{1}{2}}{\overset{2}{\int }}\frac{dx}{{\left(x-\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
${\therefore I=\frac{\pi }{4}\left(\frac{2}{\sqrt{3}}\right){\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)]}_{\frac{1}{2}}^2$
$=\frac{\pi }{2\sqrt{3}}\left(\frac{\pi }{3}-0\right)=\frac{{\pi }^2}{6\sqrt{3}}\Rightarrow A$