Question

$\int \frac{x}{x^4+x^2+1}dx$ equal to

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

Answer 1

The correct option is B $\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)$




$I=\int \frac{x}{x^4+x^2+1}dx=\int \frac{xdx}{(x^2+x+1)(x^2-x+1)}I=\frac{1}{2}\int \frac{dx}{x^2-x+1}-\frac{1}{2}\int \frac{dx}{x^2+x+1}$
$I=\frac{1}{2}\int \frac{dx}{{(x-\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}-\frac{1}{2}\frac{dx}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$
$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{x-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)-\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)$
$I=\frac{1}{\sqrt{3}}ta{n}^{-1}\left(\frac{2x^2+1}{\sqrt{3}}\right)$