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Integral

sbhatnagar

Active member
Jan 27, 2012
95
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

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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:

sbhatnagar

Active member
Jan 27, 2012
95
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!
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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