Question

\[\int\limits_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx\]

Answer 1

\[Let\, I = \int\frac{1 - x^2}{x^4 + x^2 + 1} dx\]
\[ = - \int\frac{x^2 - 1}{x^4 + x^2 + 1} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{x^2 + 1 + \frac{1}{x^2}} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{x^2 + 2 + \frac{1}{x^2} - 1} dx\]
\[ = - \int\frac{1 - \frac{1}{x^2}}{\left( x + \frac{1}{x} \right)^2 - 1} dx\]
\[Let, x + \frac{1}{x} = t\]
\[ \Rightarrow \left( 1 - \frac{1}{x^2} \right)dx = dt\]
\[\text{Then integral becomes}, \]
\[I = - \int\frac{1}{t^2 - 1} dt\]
\[ = - \frac{1}{2}\log\left| \frac{t - 1}{t + 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{t + 1}{t - 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{x + \frac{1}{x} + 1}{x + \frac{1}{x} - 1} \right|\]
\[ = \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right|\]
\[i . e . , \int\frac{1 - x^2}{x^4 + x^2 + 1} dx = \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right|\]
\[ \Rightarrow \int_0^1 \frac{1 - x^2}{x^4 + x^2 + 1} dx = \left[ \frac{1}{2}\log\left| \frac{x^2 + x + 1}{x^2 - x + 1} \right| \right]_0^1 \]
\[ = \frac{1}{2}\log 3\]