The discussion revolves around evaluating the integral \[\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}dx \quad \ x>0\] for any natural number \(m\). Participants explore various approaches to solve this integral, including substitutions and integration techniques.
While participants share similar methods and results, there is no explicit consensus on the correctness of the evaluations or the final expression derived. The discussion remains open to further verification and refinement.
Some steps in the integration process may depend on specific assumptions about the behavior of the functions involved, and the correctness of the derivative noted in the correction may affect the final result.
sbhatnagar said:Challenge Problem: For any natural number $m$, evaluate
\[\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}dx \ \quad \ x>0\]
Sudharaka said:Let, \(x=u^{\frac{1}{m}}\). Then the integral becomes,
\begin{eqnarray}
\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{\frac{1}{m}}\,dx&=&\int u(u^2+u+1)(2u^2+3u+6)^{\frac{1}{m}}\,\frac{du}{mu^{1-\frac{1}{m}}}\\
&=&\frac{1}{m}\int u^{\frac{1}{m}}(u^2+u+1)(2u^2+3u+6)^{\frac{1}{m}}du\\
&=&\frac{1}{6m}\int (2u^3+3u^2+6u)^{\frac{1}{m}}\frac{d}{du}(2u^2+3u+6)\, du\\
&=&\frac{1}{6m}\frac{(2u^3+3u^2+6u)^{\frac{1}{m}+1}}{\frac{1}{m}+1}+C\\
&=&\frac{x^{m+1}(2x^{2m}+3x^m+6)^{1+\frac{1}{m}}}{6(m+1)}+C\\
\therefore \int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{\frac{1}{m}}\,dx&=&\frac{x^{m+1}(2x^{2m}+3x^m+6)^{1+\frac{1}{m}}}{6(m+1)}+C\mbox{ for }m\in\mathbb{Z^{+}}
\end{eqnarray}
Kind Regards,
Sudharaka.
Sudharaka said:Let, \(x=u^{\frac{1}{m}}\). Then the integral becomes,
\begin{eqnarray}
\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{\frac{1}{m}}\,dx&=&\int u(u^2+u+1)(2u^2+3u+6)^{\frac{1}{m}}\,\frac{du}{mu^{1-\frac{1}{m}}}\\
&=&\frac{1}{m}\int u^{\frac{1}{m}}(u^2+u+1)(2u^2+3u+6)^{\frac{1}{m}}du\\
&=&\frac{1}{6m}\int (2u^3+3u^2+6u)^{\frac{1}{m}}\frac{d}{du}(2u^2+3u+6)\, du\\
&=&\frac{1}{6m}\frac{(2u^3+3u^2+6u)^{\frac{1}{m}+1}}{\frac{1}{m}+1}+C\\
&=&\frac{x^{m+1}(2x^{2m}+3x^m+6)^{1+\frac{1}{m}}}{6(m+1)}+C\\
\therefore \int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{\frac{1}{m}}\,dx&=&\frac{x^{m+1}(2x^{2m}+3x^m+6)^{1+\frac{1}{m}}}{6(m+1)}+C\mbox{ for }m\in\mathbb{Z^{+}}
\end{eqnarray}
Kind Regards,
Sudharaka.