Question

$\int \frac{1}{x^2-x+1}dx$

• A. $=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$
• B. $=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$
• C. $=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+C$
• D. $=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)$

Answer 1

The correct option is A $=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$

$\int \frac{1}{x^2-x+1}dx$
$=\int \frac{1}{(x-\frac{1}{2}{)}^2+\frac{3}{4}}dx$
$=\int \frac{1}{(x-\frac{1}{2}{)}^2+(\frac{\sqrt{3}}{2}{)}^2}dx$
$=\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C$
$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$