242t.08.02.09 int of rational expression.

[karush]

$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\begin{align*}
\displaystyle
I_{8.1.31}
&=\int_5^{10}
\frac{3x^5}{x^3-5} \, dx =885.576\\
\end{align*}
$\textit{ok tried getting to this answer by: }$
$u=x^3-5 \therefore du=3x^2 \, dx$
$\textit{but it didnt seem to go to good}$

 

[Prove It]

$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\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.

 

[karush]

$\tiny{242t.08.02.09}$
$\textsf{ Evaluate to nearest thousandth}\\$
\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*}
$\textit{basically}$

 

[MarkFL]

While there's technically nothing wrong with what you did, I would evaluate as follows:

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

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

 

[karush]

thanks love the tips...