Question

\[\int\frac{x^2 + 1}{x^4 + x^2 + 1} \text{  dx }\]

Answer 1

\[\text{ We have,} \]
\[I = \int \frac{\left( x^2 + 1 \right)dx}{x^4 + x^ 2 + 1}\]
\[\text{Dividing numerator and denominator by x^2 , we get}\]
\[I = \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + 1 + \frac{1}{x^2}}\]
\[ = \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{x^2 + \frac{1}{x^2} - 2 + 3}\]
\[ = \int \frac{\left( 1 + \frac{1}{x^2} \right)dx}{\left( x - \frac{1}{x} \right)^2 + 3}\]
\[\text{ Putting   x }- \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 + \frac{1}{x^2} \right)dx = dt\]
\[ \therefore I = \int \frac{dt}{t^2 + 3}\]
\[ = \int\frac{dt}{t^2 + \left( \sqrt{3} \right)^2}\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{t}{\sqrt{3}} \right) + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left[ \frac{x - \frac{1}{x}}{\sqrt{3}} \right] + C\]
\[ = \frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{x^2 - 1}{\sqrt{3} x} \right) + C\]