Question

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

• A. $\frac{4}{\sqrt{5}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{5}}\right)+C$
• B. $\frac{3}{\sqrt{2}}{\cos }^{-1}\left(\frac{x-1}{\sqrt{2}}\right)+C$
• C. $\frac{1}{\sqrt{5}}{\cos }^{-1}\left(\frac{x+1}{\sqrt{2}}\right)+C$
• D. $\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$

Answer 1

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

Let $I=\int \frac{1}{x^2-x+1}dx$
$=\frac{1}{{(x-\frac{1}{2})}^2+\frac{3}{4}}dx$
$\int \frac{1}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$
Put $x-\frac{1}{2}=t$
$\Rightarrow dx=dt$
$I=\int \frac{dt}{t^2+(\frac{\sqrt{3}}{2}{)}^2}$
$I=\frac{1}{\sqrt{3}/2}{\tan }^{-1}\left(\frac{t}{\sqrt{3}/2}\right)+C$
$=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$