MHB Integral: For any natural number , evaluate

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The integral of the expression \((x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}\) for any natural number \(m\) is evaluated using the substitution \(x=u^{1/m}\). This transforms the integral into a simpler form, leading to the result \(\frac{x^{m+1}(2x^{2m}+3x^m+6)^{1+\frac{1}{m}}}{6(m+1)} + C\). A correction was noted regarding the derivative in the integration process. The final expression holds true for \(m\) in the set of positive integers. The discussion highlights the methodical approach to solving complex integrals.
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Challenge Problem: For any natural number $m$, evaluate

\[\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}dx \ \quad \ x>0\]
 
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sbhatnagar said:
Challenge Problem: For any natural number $m$, evaluate

\[\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}dx \ \quad \ x>0\]

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^3+3u^2+6u)\, 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.
 
Last edited:
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.

Yeah, that's right!
 
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.

Typo in the third line the derivative should be of: \(2x^3+3u^2+6u\)

CB
 
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