Question

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

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

Answer 1

The correct option is B

$\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$
To solve integrals of this form we express our denominator as a perfect square,
Here, we can express the integrand as:
$\frac{1}{x^2-x+1}=\frac{1}{x^2-2\left(\frac{1}{2}\right)x+\frac{1}{4}-\frac{1}{4}+1}=\frac{1}{(x-1/2{)}^2+3/4}=\frac{1}{(x-1/2{)}^2+{(\sqrt{3}/2)}^2}$
Now, we can solve our integral as:
$I=\int \frac{\mathrm{dx}}{(x-1/2{)}^2+{(\sqrt{3}/2)}^2}\mathrm{Now},\mathrm{using}\mathrm{the}\mathrm{formula}\int \frac{\mathrm{dx}}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C\Rightarrow I=\frac{1}{\sqrt{3}/2}{\tan }^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}\right)+C\Rightarrow I=\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)+C$