Integral: For any natural number , evaluate

[sbhatnagar]

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]

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.

 

[sbhatnagar]

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]

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

 

[CellCoree]

To evaluate this integral, we can use the power rule for integration and the substitution method. First, let's rewrite the integral as:

\[\int (x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^{m}+6)^{1/m}dx = \int x^{3m+1}+x^{2m+1}+x^{m+1} \cdot (2x^{2m}+3x^{m}+6)^{1/m}dx\]

Then, we can substitute $u = 2x^{2m}+3x^{m}+6$, which means $du = (4mx^{2m-1}+3mx^{m-1})dx$. We can also rewrite the integral as:

\[\int x^{3m+1}+x^{2m+1}+x^{m+1} \cdot (2x^{2m}+3x^{m}+6)^{1/m}dx = \int x^{3m+1}+x^{2m+1}+x^{m+1} \cdot u^{1/m} \cdot \frac{(4mx^{2m-1}+3mx^{m-1})}{(4mx^{2m-1}+3mx^{m-1})} dx\]

Using the substitution $u$ and simplifying, we get:

\[\int x^{3m+1}+x^{2m+1}+x^{m+1} \cdot (2x^{2m}+3x^{m}+6)^{1/m}dx = \int \frac{x^{3m+1}+x^{2m+1}+x^{m+1} \cdot u^{1/m}}{4mx^{2m-1}+3mx^{m-1}} (4mx^{2m-1}+3mx^{m-1})dx\]

Next, we can use the power rule for integration to solve the integral inside the parentheses. This results in:

\[\int \frac{x^{3m+1}+x^{2m+1}+x^{m+1} \cdot u^{1/m}}{4mx^{2m-1}+3mx^{m-