MHB 242t.08.02.09 int of rational expression.

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The discussion focuses on evaluating the integral I_{8.1.31} = ∫_5^{10} (3x^5)/(x^3-5) dx, which results in approximately 885.576. A substitution method is attempted with u = x^3 - 5, leading to a transformed integral. The final evaluation is simplified by using the expression ∫ (1 + 5/u) du, which streamlines the calculation by reducing the number of logarithmic evaluations needed. The conclusion emphasizes that while the original method was valid, the alternative approach is more efficient for achieving the same result.
karush
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$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}
\displaystyle
I_{8.1.31}
&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx =885.576\\
\end{align*}
$\textit{ok tried getting to this answer by: }$
$u=x^3-5 \therefore du=3x^2 \, dx$
$\textit{but it didnt seem to go to good}$
 
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karush said:
$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}
\displaystyle
I_{8.1.31}
&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx =885.576\\
\end{align*}
$\textit{ok tried getting to this answer by: }$
$u=x^3-5 \therefore du=3x^2 \, dx$
$\textit{but it didnt seem to go to good}$

$\displaystyle \begin{align*} \int_5^{10}{ \frac{3\,x^5}{x^3 - 5}\,\mathrm{d}x } &= \int_5^{10}{ \frac{x^3}{x^3 - 5}\,3\,x^2\,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = x^3 - 5 \implies \mathrm{d}u = 3\,x^2\,\mathrm{d}x \end{align*}$ and note that $\displaystyle \begin{align*} u(5) = 120 \end{align*}$ and $\displaystyle \begin{align*} u(10) = 995 \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int_5^{10}{ \frac{x^3}{x^3 - 5}\,3\,x^2\,\mathrm{d}x } &= \int_{120}^{995}{ \frac{u + 5}{u}\,\mathrm{d}u } \\ &= \int_{120}^{995}{ \left( 1 + \frac{5}{u}\right) \,\mathrm{d}u } \end{align*}$

Continue.
 
$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}\displaystyle
I_{8.1.31}&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx \\
\end{align*}
$u=x^3-5 \therefore du=3x^2 \, dx$
\begin{align*}\displaystyle
I_{8.1.31}&= \int_{120}^{995}{ \frac{u + 5}{u}\,du } \\
&=\left[u+5\ln\left({\left| u \right|}\right)\right]^{995}_{120}\\
&=\left[1029.514-143.938\right] \\
&=885.576
\end{align*}
$\textit{basically}$
 
While there's technically nothing wrong with what you did, I would evaluate as follows:

$$I=\int_{120}^{995} 1+\frac{5}{u}\,du=(995-120)+5\ln\left(\frac{995}{120}\right)=5\left(175+\ln\left(\frac{199}{24}\right)\right)\approx885.576$$

This way, you only have one log to evaluate, and the rounding is done one time.
 
thanks love the tips...
 

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