Question

\[\int\frac{x^3}{x^4 + x^2 + 1}dx\]

Answer 1

\[I = \int\frac{x^3}{x^4 + x^2 + 1}dx\]
\[ = \int\frac{x^2 \cdot x}{\left( x^2 \right)^2 + x^2 + 1}dx\]
\[\text{ Let x }^2 = \text{ t or 2xdx } = dt\]
\[ \Rightarrow I = \frac{1}{2}\int\frac{t}{t^2 + t + 1}dt\]
\[ = \frac{1}{4}\int\frac{2t}{t^2 + t + 1}dt\]
\[ = \frac{1}{4}\int\frac{2t + 1 - 1}{t^2 + t + 1}dt\]

\[= \frac{1}{4}\int\left[ \frac{\left( 2t + 1 \right)}{\left( t^2 + t + 1 \right)} - \frac{1}{\left( t^2 + t + 1 \right)} \right]dt\]
\[ = \frac{1}{4}\left[ \text{ log}\left| t^2 + t + 1 \right| - \int\frac{1}{\left( t^2 + t + \frac{1}{4} + \frac{3}{4} \right)}dt \right]\]
\[ = \frac{1}{4}\left[ \text{ log }\left| t^2 + t + 1 \right| - \int\frac{1}{\left( t + \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}dt \right]\]
\[ = \frac{1}{4}\left[ \text{ log}\left| t^2 + t + 1 \right| - \frac{2}{\sqrt{3}}\tan\frac{\left( t + \frac{1}{2} \right)}{\left( \frac{\sqrt{3}}{2} \right)} \right] + c\]
\[ = \frac{1}{4}\left[ \text{ log }\left| t^2 + t + 1 \right| - \frac{2}{\sqrt{3}}\tan\left( \frac{2t + 1}{\sqrt{3}} \right) \right] + c\]

\[= \frac{1}{4}\left[ \text{ log }\left| x^4 + x^2 + 1 \right| - \frac{2}{\sqrt{3}}\tan\left( \frac{2 x^2 + 1}{\sqrt{3}} \right) \right] + c\]